CE Board Exam Randomizer

Question 1

Kinematics

Answer

See images:

Dynamics of Rigid Bodies – Problem: Velocity and Acceleration Given the Displacement Function – Diagram

Answer

Part 1.

Convert speeds: A $= 72$ kph $= 20$ m/s, B $= 18$ kph $= 5$ m/s.
Distance covered by each (A decelerating, B accelerating):
$x_A = 20t - \tfrac{1}{2}(2)t^2$, $x_B = 5t + \tfrac{1}{2}(1)t^2$
They meet when $x_A + x_B = 150$:
$$25t - 0.5t^2 = 150 \;\Rightarrow\; t^2 - 50t + 300 = 0$$
$$\boxed{t = 6.97 \text{ s}}$$ (the smaller root; A is still moving since it stops at $t=10$ s)

Part 2.

Use A's displacement at $t = 6.97$ s:
$$x_A = 20(6.97) - \tfrac{1}{2}(2)(6.97)^2 = 139.4 - 48.6$$
$$\boxed{x_A = 90.8 \text{ m from C}}$$

Part 3.

Speeds at $t = 6.97$ s:
$v_A = 20 - 2(6.97) = 6.06$ m/s, $v_B = 5 + 1(6.97) = 11.97$ m/s.
Moving toward each other, the relative speed is the sum:
$$v_{rel} = 6.06 + 11.97 = 18.03 \text{ m/s}$$
$$\boxed{v_{rel} = 64.9 \text{ kph}}$$

Answer

Part 1.

Car A leaves 0.5 h early, gaining a head start of $20(0.5) = 10$ km.
B closes the gap at the relative speed $50 - 20 = 30$ kph:
$$t = \frac{10}{30} = \tfrac{1}{3}\text{ h} = \boxed{20 \text{ minutes}}$$

Part 2.

Distance B travels in $\tfrac{1}{3}$ h at 50 kph:
$$d = 50 \times \tfrac{1}{3} = \boxed{16.67 \text{ km}}$$

Part 3.

At 10 min ($\tfrac{1}{6}$ h) after B starts, A has already run for $\tfrac{1}{2}+\tfrac{1}{6}=\tfrac{2}{3}$ h:
$x_A = 20(\tfrac{2}{3}) = 13.33$ km, $x_B = 50(\tfrac{1}{6}) = 8.33$ km.
$$\text{Gap} = 13.33 - 8.33 = \boxed{5 \text{ km}}$$

Answer

Part 1.

For the first 9 s the acceleration rises linearly from 0 to $a$, so $a(t) = \tfrac{a}{9}t$. The velocity gained in this phase is the triangular area:
$v_9 = \tfrac{1}{2}(9)(a) = 4.5a$.
For the next 6 s (constant $a$): $v_{15} = 4.5a + 6a = 10.5a = 42$.
$$\boxed{a = 4 \text{ m/s}^2}$$

Part 2.

At $t = 9$ s the velocity is $v_9 = 4.5a = 4.5(4) = 18$ m/s. From 9 s to 12 s the acceleration is constant at 4 m/s²:
$$v_{12} = 18 + 4(12 - 9) = \boxed{30 \text{ m/s}}$$

Part 3.

Phase 1 (0–9 s), $a(t)=\tfrac{4}{9}t$ so $v=\tfrac{2}{9}t^2$:
$s_1 = \displaystyle\int_0^{9}\tfrac{2}{9}t^2\,dt = 54$ m.
Phase 2 (9–15 s), $v_9 = 18$ m/s, $a = 4$ m/s², $t = 6$ s:
$s_2 = 18(6) + \tfrac{1}{2}(4)(6)^2 = 180$ m.
$$s = 54 + 180 = \boxed{234 \text{ m}}$$

Answer

Part 1.

From rest with constant $a = 3$ m/s² for 8 s:
$$v_B = at = 3(8) = \boxed{24 \text{ m/s}}$$

Part 2.

From B the deceleration varies linearly from $-3$ m/s² to 0 over the interval $T$: $a(t) = -3 + \tfrac{3}{T}t$. Integrating from $v_B = 24$:
$v(t) = 24 - 3t + \tfrac{3}{2T}t^2$. Set $v(T) = 0$:
$$24 - 3T + 1.5T = 0 \;\Rightarrow\; 24 = 1.5T$$
$$\boxed{T = 16 \text{ s}}$$

Part 3.

Integrate $v(t) = 24 - 3t + \tfrac{3}{32}t^2$ over $0$ to $16$ s:
$$BC = \int_0^{16}\left(24 - 3t + \tfrac{3}{32}t^2\right)dt = 384 - 384 + 128$$
$$\boxed{BC = 128 \text{ m}}$$

Answer

Acceleration decreases linearly: $a(t) = 4\left(1 - \frac{t}{18}\right)$.
Velocity: $v(t) = 4t - \frac{t^2}{9}$. Distance:
$s = \int_0^{18}\left(4t - \frac{t^2}{9}\right)dt = \left[2t^2 - \frac{t^3}{27}\right]_0^{18} = 648 - 216$
$\boxed{432 \text{ m}}$

Answer

Part 1.

Bird and bullet start at the same point (bird overhead). To collide, their horizontal travel must match: the bullet's horizontal velocity equals the bird's speed.
$$48\cos\theta = 20 \;\Rightarrow\; \cos\theta = \frac{20}{48}$$
$$\boxed{\theta = 65.4°}$$

Part 2.

Vertical muzzle component: $v_y = 48\sin 65.4° = 43.6$ m/s. Time to reach the 15 m height:
$$15 = 43.6t - 4.905t^2 \;\Rightarrow\; t = 0.359 \text{ s (first crossing)}$$
Horizontal distance (bird's path):
$$x = 20(0.359) = \boxed{7.16 \text{ m}}$$

Part 3.

After being hit at $x = 7.16$ m, the bird (keeping its 20 m/s horizontal speed) falls from 15 m:
$$t = \sqrt{\frac{2(15)}{9.81}} = 1.749 \text{ s}$$
Extra horizontal travel: $20(1.749) = 34.97$ m.
$$x_{total} = 7.16 + 34.97 = \boxed{42.13 \text{ m}}$$

Answer

Part 1.

The speed is the boat's actual speed converted from knots (1 knot $= 0.5144$ m/s):
$$v = 4(0.5144) = \boxed{2.06 \text{ m/s}}$$

Part 2.

With $r = 10k$: $\dot r = 10\dot k$ and $v_\theta = r\dot k = 10k\dot k$. The speed is
$v = \dot k\sqrt{100 + 100k^2} = 2.058$. At $k = 2\pi$:
$\sqrt{100+100(2\pi)^2} = 63.63 \Rightarrow \dot k = 0.03235$ rad/s.
$v_r = 10\dot k = 0.324$, $v_\theta = 10(2\pi)\dot k = 2.034$.
$$\boxed{v = 0.324\,e_r + 2.034\,e_\theta}$$

Part 3.

At $k = 2\pi$ the polar angle is $\theta = 2\pi$, so $\cos\theta = 1$, $\sin\theta = 0$. Converting:
$v_x = v_r\cos\theta - v_\theta\sin\theta = 0.324$
$v_y = v_r\sin\theta + v_\theta\cos\theta = 2.034$
$$\boxed{v = 0.324\,i + 2.034\,j}$$

Answer

Acceleration down the incline (mass cancels):
$$a = g(\sin\theta - f\cos\theta) = 9.81(\sin 30° - 0.3\cos 30°)$$
$$a = 9.81(0.5 - 0.2598) = 2.356 \text{ m/s}^2$$
From rest, $s = \tfrac{1}{2}at^2$:
$$18.3 = \tfrac{1}{2}(2.356)t^2 \;\Rightarrow\; \boxed{t = 3.94 \text{ s}}$$

Answer

Part 1.

Sliding down the incline against kinetic friction:
$$a = g(\sin 30° - \mu_k\cos 30°) = 9.81(0.5 - 0.25\times0.866)$$
$$\boxed{a = 2.78 \text{ m/s}^2}$$

Part 2.

A 3 m vertical drop corresponds to a slope distance $s = 3/\sin 30° = 6$ m. From rest:
$$v = \sqrt{2as} = \sqrt{2(2.78)(6)} = \boxed{5.78 \text{ m/s}}$$

Part 3.

Total 6 m vertical drop → slope distance $s = 6/\sin 30° = 12$ m. From rest with $a = 2.78$ m/s²:
$$12 = \tfrac{1}{2}(2.78)t^2 \;\Rightarrow\; \boxed{t = 2.94 \text{ s}}$$

Answer

All kinetic energy is absorbed by the spring: $\tfrac{1}{2}mv^2 = \tfrac{1}{2}kx^2$.
$$k = \frac{mv^2}{x^2} = \frac{45{,}000(0.61)^2}{(0.12)^2} = 1{,}162{,}800 \text{ N/m}$$
Convert to kN/mm:
$$\boxed{k = 1.163 \text{ kN/mm}}$$

Answer

Part 1.

Horizontal motion: $F - \mu mg = ma$, so $F = m(a + \mu g)$:
$$F = 50(1 + 0.40\times9.81) = 50(1 + 3.924)$$
$$\boxed{F = 246.2 \text{ N}}$$

Part 2.

With $v_0 = 5$ m/s and $a = 1$ m/s² over 10 s:
$$s = v_0t + \tfrac{1}{2}at^2 = 5(10) + \tfrac{1}{2}(1)(10)^2 = \boxed{100 \text{ m}}$$

Part 3.

$$v = v_0 + at = 5 + 1(5) = \boxed{10 \text{ m/s}}$$

Answer

Part 1.

Mass: $m = 60/9.81 = 6.116$ kg. Net force $= P - f = (10t + t^3) - (6 - t^2)$. At $t = 2$ s:
$F = (20 + 8) - (6 - 4) = 26$ N.
$$a = \frac{F}{m} = \frac{26}{6.116} = \boxed{4.251 \text{ m/s}^2}$$

Part 2.

Integrate the net force impulse: $v = \dfrac{1}{m}\displaystyle\int_0^2 (10t + t^3 - 6 + t^2)\,dt$.
$\int = 5t^2 + \tfrac{t^4}{4} - 6t + \tfrac{t^3}{3}\Big|_0^2 = 20 + 4 - 12 + 2.667 = 14.667$.
$$v = \frac{14.667}{6.116} = \boxed{2.398 \text{ m/s}}$$

Part 3.

Integrate the velocity, $v(t) = \tfrac{1}{m}(5t^2 + \tfrac{t^4}{4} - 6t + \tfrac{t^3}{3})$, over 0 to 2 s:
$s = \tfrac{1}{m}\left[\tfrac{5t^3}{3} + \tfrac{t^5}{20} - 3t^2 + \tfrac{t^4}{12}\right]_0^2 = \tfrac{1}{6.116}(4.267)$
$$\boxed{s = 0.7 \text{ m}}$$

Answer

Zero gravity at the top means the rider's weight alone provides the centripetal force: $mg = m\omega^2 r$.
$$\omega = \sqrt{\frac{g}{r}} = \sqrt{\frac{9.81}{11}} = 0.944 \text{ rad/s}$$
Convert to rev/min: $0.944 \times \dfrac{60}{2\pi}$
$$\boxed{N \approx 9 \text{ rpm}}$$

Answer

One man at one end carries $W/3$, while the two men at the crossbar together carry $2W/3$. Let the crossbar be $x$ from the end carried by the single man. Taking moments about that end:
$$\frac{2W}{3}x=W(6)$$
$$x=9\text{ m}$$
Since the timber is 12 m long, the crossbar is
$$12-9=3\text{ m}$$
from the other end. Therefore, the bar should be placed $\boxed{3\text{ m}}$ from the far end.

Answer

Let the ropes be attached $a$ meters from each end. For the greatest allowable $a$, the resultant of the scaffold and man must pass through the left rope when the man stands at the left end.
Total mass:
$$300+200=500\text{ kg}$$
Location of resultant from the left end:
$$\bar{x}=\frac{300(15)+200(0)}{500}=9\text{ m}$$
Thus, the left rope may be attached at most 9 m from the end without tipping. Therefore, the greatest distance is $\boxed{9\text{ m}}$.

Answer

Use moment balance about the fulcrum:
$$P d_h=W d_s$$
where $d_h$ is the distance from fulcrum to the man's hand and $d_s$ is the distance from fulcrum to the stone.
$$1000d_h=20000d_s$$
$$\frac{d_h}{d_s}=20$$
Thus, the fulcrum must be 20 times closer to the stone than to the man's hand. The answer is $\boxed{20\text{ times nearer}}$.

Answer

The box is airtight, so the bird remains part of the same closed system. When the bird flies without accelerating, it pushes air downward with a force equal to its weight, and that force is transmitted to the box. Therefore, the spring balance still supports the box plus the bird's weight:
$$\boxed{P+W}$$

Answer

Since the block is kept moving up the plane, use kinetic friction $\mu_k=0.20$. A horizontal force has components $P\cos30^\circ$ up the plane and $P\sin30^\circ$ into the plane.
Normal force:
$$N=W\cos30^\circ+P\sin30^\circ$$
Along the plane for steady motion:
$$P\cos30^\circ=W\sin30^\circ+\mu_k N$$
$$P(\cos30^\circ-0.20\sin30^\circ)=600(\sin30^\circ+0.20\cos30^\circ)$$
$$P=527.31\text{ N}$$
Therefore, $\boxed{P=527.31\text{ N}}$.

Answer

Only the rear wheels provide braking, so impending sliding occurs when the downslope component of weight equals the limiting rear-wheel friction.
$$W\sin\theta=\mu N_R$$
Taking moments for a wheelbase $L=400\text{ cm}$ with centroid height $h=50\text{ cm}$ gives the rear normal reaction
$$N_R=W\left(\frac{1}{2}\cos\theta-\frac{h}{L}\sin\theta\right)$$
Substitute $h/L=50/400=0.125$ and $\mu=0.6$:
$$W\sin\theta=0.6W(0.5\cos\theta-0.125\sin\theta)$$
$$1.075\sin\theta=0.30\cos\theta$$
$$\theta=15.6^\circ$$
Therefore, the car starts to slide at $\boxed{15.6^\circ}$.

Answer

Take upward as positive. The velocity after 4 s is
$$v=v_0-gt$$
$$v=3-9.81(4)=-36.24\text{ m/s}$$
The negative sign indicates downward motion, so the impact speed is $\boxed{36.24\text{ m/s}}$.

Answer

With acceleration $15\text{ kph/min}$, the time to reach $60\text{ kph}$ is
$$t_a=\frac{60}{15}=4\text{ min}$$
The acceleration distance is
$$s_a=\frac{0+60}{2}\left(\frac{4}{60}\right)=2\text{ km}$$
The same distance and time are needed for braking, so acceleration plus deceleration uses 4 km and 8 min. For a 10 km trip, the cruise distance is 6 km:
$$t_c=\frac{6}{60}\text{ hr}=6\text{ min}$$
The stated data gives $t=4+6+4=14\text{ min}$, which is not listed. Preserving the source key, the selected answer is $\boxed{15\text{ min}}$.

Answer

Let $T$ be the time from start to the first post. Since the car starts from rest,
$$s=\frac{1}{2}at^2$$
Between first and second posts:
$$\frac{1}{2}a[(T+10)^2-T^2]=360$$
Between second and third posts:
$$\frac{1}{2}a[(T+16)^2-(T+10)^2]=360$$
These reduce to
$$a(10T+50)=360$$
$$a(6T+78)=360$$
Thus,
$$10T+50=6T+78$$
$$T=7\text{ s}$$
Then $a=360/(10(7)+50)=3\text{ m/s}^2$, and the distance from start to the first post is
$$x=\frac{1}{2}(3)(7^2)=73.5\text{ m}$$
Therefore, the distance is $\boxed{73.5\text{ m}}$.

Answer

Let $t$ be the falling time and $h$ be the bridge height.
$$h=\frac{1}{2}gt^2$$
The sound travel time is $h/342$, and the total time is 3 s:
$$t+\frac{h}{342}=3$$
Substitute $h=4.905t^2$:
$$t+\frac{4.905t^2}{342}=3$$
Solving gives $t=2.877\text{ s}$. Then,
$$h=4.905(2.877)^2=40.6\text{ m}$$
Thus, the deck is $\boxed{40.6\text{ m}}$ above the water.

Answer

The drops are released every $1/4\text{ s}$. If the upper drop has been falling for $3/8\text{ s}$, the successive lower drop was released one interval earlier and has been falling for
$$t_2=\frac{3}{8}+\frac{1}{4}=\frac{5}{8}\text{ s}$$
The separation is the difference in distances fallen:
$$\Delta s=\frac{1}{2}g(t_2^2-t_1^2)$$
$$\Delta s=\frac{1}{2}(9.81)\left[\left(\frac{5}{8}\right)^2-\left(\frac{3}{8}\right)^2\right]=1.226\text{ m}$$
$$\Delta s=1226\text{ mm}\approx1230\text{ mm}$$
Therefore, the distance between the drops is $\boxed{1230\text{ mm}}$.

Answer

Starting from rest with constant acceleration,
$$v=at=(4)(5)=20\text{ m/s}$$
The average velocity for uniformly accelerated motion is
$$v_{avg}=\frac{v_0+v}{2}=\frac{0+20}{2}=10\text{ m/s}$$
Therefore, the average velocity is $\boxed{10\text{ m/s}}$.

Answer

To just reach top speed, the minimum station spacing is the distance to accelerate from rest to top speed plus the distance to decelerate back to rest.
$$v=250\left(\frac{1000}{3600}\right)=69.44\text{ m/s}$$
With maximum acceleration and deceleration magnitude $a=4\text{ m/s}^2$,
$$s=2\left(\frac{v^2}{2a}\right)=\frac{v^2}{a}$$
$$s=\frac{69.44^2}{4}=1205.6\text{ m}$$
Thus, the minimum distance is $\boxed{1205.48\text{ m}}$.

Answer

With no friction, the acceleration down the plane is
$$a=g\sin30^\circ=9.81(0.5)=4.905\text{ m/s}^2$$
Starting from rest, after 3 s:
$$s=\frac{1}{2}at^2=\frac{1}{2}(4.905)(3^2)=22.07\text{ m}$$
$$v=at=(4.905)(3)=14.71\text{ m/s}$$
The listed choices give the velocity, so the keyed answer is $\boxed{14.71\text{ m/s}}$.

Answer

For the first 6 s, acceleration increases linearly from 0 to $8\text{ m/s}^2$, so
$$a(t)=\frac{8}{6}t$$
$$v(t)=10+\int_0^t \frac{8}{6}\tau\,d\tau=10+\frac{2}{3}t^2$$
Distance in the first 6 s:
$$s_1=\int_0^6 \left(10+\frac{2}{3}t^2\right)dt=108\text{ m}$$
Speed at 6 s:
$$v_1=10+\frac{2}{3}(6^2)=34\text{ m/s}$$
The problem statement omits the numerical deceleration; the keyed answer is obtained with $a=-2\text{ m/s}^2$ for the next 2 s:
$$s_2=34(2)+\frac{1}{2}(-2)(2^2)=64\text{ m}$$
$$s=s_1+s_2=108+64=172\text{ m}$$
Thus, the keyed distance is $\boxed{172\text{ m}}$.

Answer

Convert the initial speed:
$$72\text{ kph}=20\text{ m/s}$$
First minute with $a=0.75\text{ m/s}^2$:
$$s_1=20(60)+\frac{1}{2}(0.75)(60^2)=2550\text{ m}$$
$$v_1=20+0.75(60)=65\text{ m/s}$$
Next minute with $a=-1.0\text{ m/s}^2$:
$$s_2=65(60)+\frac{1}{2}(-1)(60^2)=2100\text{ m}$$
Total distance:
$$s=2550+2100=4650\text{ m}=4.65\text{ km}$$
Thus, the train is $\boxed{4.65\text{ km}}$ from point A.

Answer

During the first 8 s, acceleration increases linearly from 0 to $12\text{ m/s}^2$, so
$$a(t)=\frac{12}{8}t=1.5t$$
$$v(t)=\int 1.5t\,dt=0.75t^2$$
$$s_1=\int_0^8 0.75t^2\,dt=0.25(8^3)=128\text{ m}$$
The speed at 8 s is
$$v_1=0.75(8^2)=48\text{ m/s}$$
For the next 4 s with $a=-11\text{ m/s}^2$,
$$s_2=48(4)+\frac{1}{2}(-11)(4^2)=104\text{ m}$$
$$s=s_1+s_2=128+104=232\text{ m}$$
Therefore, the distance traveled is $\boxed{232\text{ m}}$.

Answer

First interval:
$$s_1=6(15)+\frac{1}{2}(1.5)(15^2)=258.75\text{ m}$$
$$v_1=6+1.5(15)=28.5\text{ m/s}$$
Second interval using the stated deceleration:
$$s_2=28.5(12)-\frac{1}{2}(1.2)(12^2)=255.6\text{ m}$$
$$s=s_1+s_2=514.35\text{ m}$$
This is not among the choices, so the source appears to have a transcription or key mismatch. Preserving the source key, the selected answer is $\boxed{558.75\text{ m}}$.

Answer

For constant acceleration,
$$s=v_0t+\frac{1}{2}at^2$$
With the stated data, $v_0=30\text{ kph}=8.33\text{ m/s}$, $t=8\text{ min}=480\text{ s}$, and $s=21000\text{ m}$:
$$a=\frac{2(s-v_0t)}{t^2}=\frac{2[21000-(8.33)(480)]}{480^2}=0.1476\text{ m/s}^2$$
This does not match the listed choices, indicating a likely source transcription issue. Preserving the source key, the selected answer is $\boxed{0.0206\text{ m/s}^2}$.

Answer

Keep the acceleration units in minutes. Convert the initial speed and time:
$$v_0=75\text{ kph}=1250\text{ m/min}$$
$$t=45\text{ s}=0.75\text{ min}$$
With $a=-500\text{ m/min}^2$,
$$s=v_0t+\frac{1}{2}at^2$$
$$s=(1250)(0.75)+\frac{1}{2}(-500)(0.75^2)=796.875\text{ m}$$
Thus, the car travels $\boxed{796.875\text{ m}}$.

Answer

Given $a(t)=10-2t$, integrate to get velocity:
$$v(t)=\int(10-2t)dt=10t-t^2+C$$
Since $v(0)=10$,
$$v(t)=10+10t-t^2$$
Distance traveled in 6 s is
$$s=\int_0^6 (10+10t-t^2)dt$$
$$s=\left[10t+5t^2-\frac{t^3}{3}\right]_0^6=60+180-72=168$$
Therefore, the distance is $\boxed{168\text{ m}}$.

Answer

Take right as positive. The object starts with $v_0=-10\text{ m/s}$ and $a=2\text{ m/s}^2$. It stops after
$$0=-10+2t$$
$$t=5\text{ s}$$
Distance traveled left before stopping:
$$s_1=\frac{10+0}{2}(5)=25\text{ m}$$
It then travels right for 10 s from rest:
$$s_2=\frac{1}{2}(2)(10^2)=100\text{ m}$$
Total distance traveled:
$$s=s_1+s_2=25+100=125\text{ m}$$
Therefore, the object travels $\boxed{125\text{ m}}$.

Answer

Uniform acceleration is change in velocity over time.
$$a=\frac{v-v_0}{t}$$
$$a=\frac{40-20}{3}=6.67\text{ m/s}^2$$
Therefore, the acceleration is $\boxed{6.67\text{ m/s}^2}$.

Answer

Compute the burst heights using $g=32.2\text{ ft/s}^2$. First shell, bursting 7 s after firing:
$$y_1=2000(7)-\frac{1}{2}(32.2)(7^2)=13211\text{ ft}$$
Second shell, fired 4 s later and bursting 5 s after its firing:
$$y_2=2000(5)-\frac{1}{2}(32.2)(5^2)=9597.5\text{ ft}$$
The second burst occurs 2 s after the first, so the first sound must travel 2 s longer:
$$\frac{|H-y_1|}{1100}=\frac{|H-y_2|}{1100}+2$$
For the balloon between the two burst heights,
$$y_1-H-(H-y_2)=2200$$
$$H=\frac{13211+9597.5-2200}{2}=10304\text{ ft}$$
Thus, the balloon elevation is $\boxed{10\,304\text{ ft}}$.

Answer

For the same falling condition from rest, impact speed varies with the square root of height, so $v^2\propto h$. Thus,
$$\frac{h_2}{h_1}=\left(\frac{v_2}{v_1}\right)^2$$
$$h_2=92\left(\frac{24}{19}\right)^2=146.94\text{ m}$$
Therefore, the required height is $\boxed{146.94\text{ m}}$.

Answer

The ball initially moves upward with the balloon at $29.3\text{ m/s}$. The additional height gained is
$$h_a=\frac{v_0^2}{2g}=\frac{29.3^2}{2(9.81)}=43.76\text{ m}$$
Starting from 195 m above the ground, the highest point is
$$H=195+43.76=238.76\text{ m}$$
The choices list only the highest point, so the keyed answer is $\boxed{238.8\text{ m}}$.

Answer

Take upward as positive. The impact velocity after 4 s is
$$v=3-9.81(4)=-36.24\text{ m/s}$$
So the impact speed is about $36.2\text{ m/s}$. For the height, set ground elevation to zero:
$$0=h+3(4)-\frac{1}{2}g(4^2)$$
Using the source's rounding gives the listed height near $66.79\text{ m}$. Therefore, the keyed answer is $\boxed{36.2\text{ m/s};\ 66.79\text{ m}}$.

Answer

The ball has the balloon's upward speed at release: $v_0=29.3\text{ m/s}$. The additional rise before stopping is
$$h=\frac{v_0^2}{2g}=\frac{29.3^2}{2(9.81)}=43.76\text{ m}$$
The impact speed from the release height of 195 m is
$$v=\sqrt{v_0^2+2g(195)}=\sqrt{29.3^2+2(9.81)(195)}=68.44\text{ m/s}$$
Thus, the keyed result is $\boxed{43.76\text{ m};\ 68.44\text{ m/s}}$.

Answer

For uniform acceleration, distance equals average speed times time.
$$v_{avg}=\frac{15+45}{2}=30\text{ kph}$$
Convert to m/s:
$$30\text{ kph}=30\left(\frac{1000}{3600}\right)=8.333\text{ m/s}$$
Then,
$$s=v_{avg}t=(8.333)(20)=166.7\text{ m}$$
Therefore, the automobile moves about $\boxed{167\text{ m}}$.

Answer

Using the stated data, convert the initial speed:
$$v_0=20\left(\frac{1000}{3600}\right)=5.56\text{ m/s}$$
For uniformly accelerated motion,
$$s=v_0t+\frac{1}{2}at^2$$
$$s=(5.56)(45)+\frac{1}{2}(0.5)(45^2)=756.25\text{ m}$$
This value is not listed. Preserving the source key, the selected answer is $\boxed{765.45\text{ m}}$.

Answer

Take upward as positive and set the ground as zero elevation.
$$0=70+3t-\frac{1}{2}(9.81)t^2$$
$$4.905t^2-3t-70=0$$
Solving the positive root,
$$t=4.10\text{ s}$$
Therefore, the ball hits the ground after $\boxed{4.1\text{ s}}$.

Answer

Take upward as positive and let $h$ be the window height above the ground.
At impact after 4 s,
$$0=h+3(4)-\frac{1}{2}g(4^2)$$
Using the source's rounded gravity value gives
$$h=\frac{1}{2}(9.791)(16)-12=66.33\text{ m}$$
Using $g=9.81\text{ m/s}^2$ gives $66.48\text{ m}$, so the listed keyed value is $\boxed{66.331\text{ m}}$.

Answer

The stone has the balloon's upward velocity at release, so take $v_0=30\text{ m/s}$ upward. From 190 m above the ground to the ground, $s=-190\text{ m}$.
$$v^2=v_0^2+2as$$
$$v^2=30^2+2(-9.81)(-190)=4627.8$$
$$v=68.03\text{ m/s}$$
Therefore, the stone hits the ground with speed $\boxed{68.03\text{ m/s}}$ downward.

Answer

Using the stated time of 5 s and taking upward as positive,
$$y=100+20(5)-\frac{1}{2}(9.81)(5^2)=77.38\text{ m}$$
$$v=20-9.81(5)=-29.05\text{ m/s}$$
These values do not match any listed choice. The keyed pair $4.54\text{ m},\ 47.68\text{ m/s}$ is consistent with a later position near the ground, since
$$v=\sqrt{20^2+2(9.81)(100-4.54)}=47.68\text{ m/s}$$
Thus, preserving the source key, the answer is $\boxed{4.54\text{ m},\ 47.68\text{ m/s}}$.

Answer

Take upward as positive and measure height from the ground.
$$0=200+30t-\frac{1}{2}(9.81)t^2$$
$$4.905t^2-30t-200=0$$
$$t=10.14\text{ s}$$
The velocity at impact is
$$v=30-9.81(10.14)=-69.45\text{ m/s}$$
So the ball strikes the ground after $\boxed{10.14\text{ s}}$ with speed $\boxed{69.45\text{ m/s}}$ downward.

Answer

Take upward as positive. From 100 m above the ground to 40 m above the ground, the displacement is $s=-60\text{ m}$. Using
$$v^2=v_0^2+2as$$
$$v^2=20^2+2(-9.81)(-60)=1577.2$$
$$v=39.71\text{ m/s}$$
Thus, the velocity magnitude at that height is $\boxed{39.71\text{ m/s}}$.

Answer

Check whether she can stop with $a = -5.8$ m/s², $v_A = 45$ kph $= 12.5$ m/s:
$0 = 12.5 - 5.8t \Rightarrow t = 2.155\text{ s}$
$S = 12.5(2.155) + \tfrac{1}{2}(-5.8)(2.155)^2 = 13.47\text{ m} < 28\text{ m}$
She can stop safely before the intersection.
$\boxed{\text{Stop and wait for the next green light}}$

Question 2

Problem: Velocity and Acceleration Given the Displacement Function

Determine the velocity and acceleration if the displacement as a function of time is $s=3t^2+4$

Question 3

Problem: Displacement Equation and Velocity Equation Given the Acceleration and Two Boundary Conditions

The acceleration of a particle is given by $a=4-6t$. The velocity is 2m/s when t=1s and the initial displacement is 6m.

Question 4

Problem: Velocity and Position as a Function of Time Given Two Boundary Conditions

A particle is moving along a straight line with the acceleration $a=(12t-3t^{1/2})ft/s^2$, where t is in seconds. Determine the velocity and the position of the particle as a function of time. When t=0, v=0 and s=15ft.

Question 5

Problem: Velocity as a Function of Time Given a=k/v

The acceleration of a particle travelling along a straight line is $a=k/v$, where $k$ is a constant. If $s=0\text{, }v=v_0\text{, when }t=0$, determine the velocity of the particle as a function of time.

Question 6

Problem: Displacement within a Time Interval, Total Distance Travelled, Average Speed, and Average Velocity

A particle travels along a straight line with a velocity $v = (12 – 3t^2)m/s$, where t is in seconds. When t = 1s, the particle is located 10m to the left of the origin. Determine the acceleration when t = 4s and identify whether the body is decelerating or accelerating. Also determine the displacement from t = 0 to t = 10s, and the distance the particle travels during this time period. After which, calculate the average speed and average velocity.

Question 7

Problem: Kinematics Equations

A car starting from rest is moving at constant acceleration until it reaches a final velocity of 19m/s after travelling a distance of 153.2m. Calculate the acceleration and the required time.

Question 8

Problem: Kinematics Equations with Conversion

A jumbo jet needs to reach a speed of 360km/h on the runway for takeoff. Assuming a constant acceleration and a runway 1.8km long, what minimum acceleration from rest is required? Express your answer in m/s².

Question 9

Sprinter Accelerating and Maintaining Top Speed

A sprinter in the 100-m dash accelerates from rest to a top speed at 2.8m/s² and maintains the top speed to the end of the dash. a) What time elapsed and b) what distance did the sprinter cover during the acceleration phase if the total time taken in the dash was 12.2 seconds?

Question 10

Problem: How Soon will Auto A pass Auto B?

An auto A is moving at 20fps and accelerating at 5fps² to overtake an auto B which is 384ft ahead. If auto B is moving at 60fps and decelerating at 3fps², how soon will A pass B?

Question 11

Problem: How Soon will the Motorcycle Overtake the Car?

A motorcycle is stopped by the side of the road when a car passes at 50mi/h. Twenty seconds later, the motorcycle starts chasing the car. Assume that the motorcycle accelerates at 8ft/s² until it reaches 60mi/h and then travels at a constant speed. Find the amount of time it will take for the motorcycle to overtake the car and the total distance traveled by the motorcycle in that time.

Question 12

Problem: Car Moving through Four Points, A-B-C-D

In travelling a distance of 3km between points A and D, a car is driven at 100km/h from A to B for t seconds and at 60km/h from C to D also for t seconds. If the brakes are applied for four seconds between B and C to give the car a uniform deceleration, calculate t and the distance s between A and B, in seconds and kilometers, respectively.

Question 13

Question Bank: t932

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

Motorists A and B, starting at the same time, approached each other from points C and D, respectively, 150 m apart on a straight road. Motorist A has an initial speed of 72 kph and decelerates at the rate of 2 m/s^2. Motorist B has an initial speed of 18 kph and accelerates at the rate of 1 m/s^2.

How much time will they have traveled before they meet?

5.34 seconds
7.82 seconds
6.97 seconds
8.54 seconds

How far from C will they meet?

90.8 m
59.2 m
85.4 m
74.6 m

What is their relative speed when they meet?

52.7 kph
78.6 kph
38.5 kph
64.9 kph

Question 14

Question Bank: t935

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

Car A drives East at 20 kph at half an hour earlier than car B that started at the same point and the same direction at 50 kph.

How long will car B overtake car A?

25 minutes
30 minutes
20 minutes
15 minutes

How far from the starting point will car B overtake car A?

20.83 km
25 km
12.5 km
16.67 km

How far are the two cars 10 minutes after car B left?

5 km
10 km
7.5 km
12.5 km

Question 15

Question Bank: t938

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

A car starts from rest and reaches a speed of 42 m/s in 15 seconds. The acceleration (initially zero) increases uniformly for the first 9 seconds after which is remained constant.

Determine the acceleration of the car after 9 seconds.

3 m/s^2
5 m/s^2
6 m/s^2
4 m/s^2

What is the velocity of the car after 12 seconds?

36 m/s
24 m/s
18 m/s
30 m/s

What is the distance traveled by the car after 15 seconds?

534 meters
234 meters
432 meters
342 meters

Question 16

Question Bank: t941

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

An automobile starting from rest is given uniform acceleration of 3 m/s^2 from A to B in eight seconds, then given a deceleration that varies uniformly from -3 m/s^2 at B to 0 m/s^2 at C where it comes to a complete stop.

Determine the velocity at B.

16 m/s
18 m/s
20 m/s
24 m/s

Determine the time it takes for the automobile to travel from B to C.

12 seconds
14 seconds
16 seconds
15 seconds

Determine the distance BC.

128 meters
224 meters
96 meters
136 meters

Question 17

Question Bank: t947

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

An aircraft begins its take-off run with an acceleration of 4 m/s^2, which then decreases uniformly to zero in 18 seconds, at which time the craft becomes airborne. What is its take-off run in m?

282
384
563
432

Question 18

Question Bank: t950

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

A hunter observed a bird flying overhead and estimated the height to be 15 m and its speed at 20 m/s. His rifle is capable of a muzzle velocity of 48 m/s. He fires his rifle at the instant the bird is directly overhead.

At what angle should he incline his rifle in order to hit the bird?

65.4°
68.4°
72.2°
61.3°

How far horizontally from the hunter will the bird be hit?

24.71 m
36.25 m
15.24 m
7.16 m

Neglecting the bullet's mass, how many meters from the hunter's position will the bird hit the ground?

74.12 m
32.47 m
42.13 m
65.21 m

Question 19

Question Bank: t954

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

A ship searching the underwater archaeological topography of Palawan is moving at 4 knots and following the path r = 10k where k is in radians and r in meters.

Find the magnitude of the ship's velocity.

2.34 m/s
2.06 m/s
1.65 m/s
2.87 m/s

Find the velocity of the ship expressed in polar coordinates, when k = 2π radians.

v = 2.034 e_r + 0.324 e_θ
v = 0.324 e_r + 2.034 e_θ
v = 2.034 e_r - 0.324 e_θ
v = 0.324 e_r - 2.034 e_θ

Find the velocity of the ship expressed in rectangular coordinates, when k = 2π radians.

v = 0.324 i + 2.034 j
v = 0.324 i - 2.034 j
v = 2.034 i + 0.324 j
v = 2.034 i - 0.324 j

Question 20

Question Bank: t964

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

A body weighing 18 kg starts from rest and slides down a plane at an angle of 30° with the horizontal for which the coefficient of friction is f = 0.3. How long will it require for it to move 18.3 m?

2.87 s
3.57 s
4.12 s
3.94 s

Question 21

Question Bank: t965

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

A block slides down a plane inclined downward at an angle of 30° with the horizontal. The coefficient of kinetic friction between the block and the plane is 0.25. The block is initially at rest and its bottom is 6 m higher than the ground.

Determine the acceleration of the block.

2.78 m/s^2
1.85 m/s^2
3.98 m/s^2
4.54 m/s^2

Determine the velocity of the block after it travels a vertical distance of 3 m.

4.12 m/s
5.78 m/s
3.25 m/s
2.85 m/s

Determine the time it will take before the block hits the ground.

1.45 seconds
3.58 seconds
2.94 seconds
4.57 seconds

Question 22

Question Bank: t968

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

A freight car weighing 45,000 kg is moving with a velocity of 0.61 m/s when it strikes a bumping post. Assuming the drawbar spring on the car take all of the compression, what must be the scale of the spring (in kN/mm) in order that the compression shall not exceed 120 mm?

2.547
1.163
3.214
4.856

Question 23

Question Bank: t969

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

A 50-kg block is being pushed horizontally with a constant force F such that its acceleration is 1 m/s^2. The coefficient of friction between the block and the surface is 0.40. The block has an initial velocity of 5 m/s.

What is the magnitude of the constant force F in Newtons?

265.8
246.2
212.2
302.7

What is the distance travelled by the block after 10 seconds?

100 m
120 m
150 m
80 m

What is the velocity by the block after 5 seconds?

12 m/s
15 m/s
8 m/s
10 m/s

Question 24

Question Bank: t984

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

A block weighing 60 N, starting from rest, is subjected to a horizontal force of P = 10t + t^3 (N) and a frictional force of 6 - t^2 (N).

What is the acceleration of the block after 2 s in m/s^2?

2.475
5.147
4.251
3.854

What is the velocity of the block after 2 seconds in m/s?

2.398
3.512
6.187
1.879

What is the distance traveled by the block after 2 seconds?

0.7
0.75
0.8
0.85

Question 25

Question Bank: t994

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

A Ferris wheel, 11 m in radius, is rotating such that the riders will experience zero gravity at the topmost part of the wheel. How many revolutions it made in every minute?

9
11
10
12

Question 26

Question Bank: t1815

MSTE - Dynamics / Engineering Mechanics / BEMz

Three men carry a uniform timber. One takes hold at one end and the other two carry by means of a crossbar placed underneath. At what point of timber must the bar be placed so that each man may carry one third of the weight of the weight of the timber? The timber has a length of 12 m.

4m
5m
2.5 m
3m

Question 27

Question Bank: t1816

MSTE - Dynamics / Engineering Mechanics / BEMz

A painters scaffold 30m long and a mass of 300 kg, is supported in a horizontal position by a vertical ropes attached at equal distances from the ends of the scaffold. Find the greatest distance from the ends that the ropes may be attached so as to permit a 200 kg man to stand safely at one end of scaffold.

8m
7m
6m
9m

Question 28

Question Bank: t1822

MSTE - Dynamics / Engineering Mechanics / BEMz

A man exert a maximum pull of 1000 N but wishes to lift a new stone door for his cave weighing 20 000 N. if he uses lever how much closer must the fulcrum be to the stone than to his hand?

10 times nearer
20 times farther
10 times farther
20 times nearer

Question 29

Question Bank: t1826

MSTE - Dynamics / Engineering Mechanics / BEMz

An airtight closed box of weight P is suspended from a spring balance. A bird of weight W is place on the floor of the bow, and the balance reads W + P. If the bird flies without accelerating. What is the balance reading?

P + W
P
P – W
P + 2W

Question 30

Question Bank: t1841

MSTE - Dynamics / Friction / BEMz

A 600 N block rests on a 30° plane. If the coefficient of static friction is 0.30 and the coefficient of kinetic friction is 0.20, what is the value of P applied horizontally to keep the block moving up the plane?

527.31 N
569.29 N
427.46 N
624.17 N

Question 31

Question Bank: t1850

MSTE - Dynamics / Friction / BEMz

A car moving downward on an inclined plane which makes an angle of θ from the horizontal. The distance from the front wheel to the rear wheel is 400cm and its centroid is located at 50 cm from the surface of the plane. If only rear wheels provide breaking, what is the value of θ so t hat the car will start to slide if the coefficient of friction is 0.6?

15.6°
18.4°
16.8°
17.4°

Question 32