Question 1

figure is a graph of $\mathrm{v}_{\mathrm{x}}(\mathrm{t})$ for a car. Solve graphically for the distance traveled from $\mathrm{t}=10 \mathrm{~s}$ to $\mathrm{t}=15 \mathrm{~s}$.&#10; &#10;A) $75 \mathrm{~m}$&#10;B) $69 \mathrm{~m}$&#10;C) $67 \mathrm{~m}$&#10;D) $70 \mathrm{~m}$

Accepted Answer

$69 \mathrm{~m}$&#10;

Question 2

figure shows the graph of $v_{x}$ versus time for an object moving along the $x$-axis. Solve graphically for the distance traveled from $\mathrm{t}=9.0 \mathrm{~s}$ to $\mathrm{t}=13.0 \mathrm{~s}$.&#10; &#10;A) $84 \mathrm{~m}$&#10;B) $80 \mathrm{~m}$&#10;C) $60 \mathrm{~m}$&#10;D) $76 \mathrm{~m}$

Accepted Answer

$80 \mathrm{~m}$&#10;

Question 3

car travels a distance of $100 \mathrm{~km}$ in 2.00 hours. It then travels an additional distance of $60.0 \mathrm{~km}$ in 1.00 hour. The average speed of the car for the entire trip is&#10;A) $50.0 \mathrm{~km} / \mathrm{hr}$.&#10;B) $46.7 \mathrm{~km} / \mathrm{hr}$.&#10;C) $80.0 \mathrm{~km} / \mathrm{hr}$.&#10;D) $60.0 \mathrm{~km} / \mathrm{hr}$.&#10;E) $53.3 \mathrm{~km} / \mathrm{hr}$.

Accepted Answer

The total distance traveled is $100 \mathrm{~km} + 60.0 \mathrm{~km} = 160.0 \mathrm{~km}$, and the total time is $2.00 \mathrm{~hr} + 1.00 \mathrm{~hr} = 3.00 \mathrm{~hr}$. The average speed is $160.0 \mathrm{~km} / 3.00 \mathrm{~hr} = 53.3 \mathrm{~km/hr}$.

Question 4

car travels at $50.0 \mathrm{~km} / \mathrm{hr}$ for 2.00 hours. It then travels an additional distance of $40.0 \mathrm{~km}$ in 1.00 hour. The average speed of the car for the entire trip is&#10;A) $30.0 \mathrm{~km} / \mathrm{hr}$.&#10;B) $53.3 \mathrm{~km} / \mathrm{hr}$.&#10;C) $57.1 \mathrm{~km} / \mathrm{hr}$.&#10;D) $61.0 \mathrm{~km} / \mathrm{hr}$.&#10;E) $46.7 \mathrm{~km} / \mathrm{hr}$.

Accepted Answer

The car travels $100.0 \mathrm{~km}$ in the first 2 hours ($50.0 \mathrm{~km/hr} \times 2.00 \mathrm{~hr}$) and $40.0 \mathrm{~km}$ in the next hour, totaling $140.0 \mathrm{~km}$ over 3 hours. The average speed is $140.0 \mathrm{~km} / 3 \mathrm{~hr} = 46.7 \mathrm{~km/hr}$.

Question 5

car travels a distance of $140 \mathrm{~km}$ at $70.0 \mathrm{~km} / \mathrm{hr}$ . It then travels an additional distance of $60.0 \mathrm{~km}$ at 40.0 $\mathrm{km} / \mathrm{hr}$ . The average speed is

A)

57.1 \mathrm{~km} / \mathrm{hr}

.
B)

46.7 \mathrm{~km} / \mathrm{hr}

.
C)

45.0 \mathrm{~km} / \mathrm{hr}

.
D)

61.0 \mathrm{~km} / \mathrm{hr}

.
E)

53.3 \mathrm{~km} / \mathrm{hr}

.

Accepted Answer

The average speed is calculated by dividing the total distance by the total time. The total distance is $140 \mathrm{~km} + 60.0 \mathrm{~km} = 200 \mathrm{~km}$. The time for the first part is $140 \mathrm{~km} / 70.0 \mathrm{~km/hr} = 2 \mathrm{~hr}$ and for the second part is $60.0 \mathrm{~km} / 40.0 \mathrm{~km/hr} = 1.5 \mathrm{~hr}$. The total time is $2 \mathrm{~hr} + 1.5 \mathrm{~hr} = 3.5 \mathrm{~hr}$. Therefore, the average speed is $200 \mathrm{~km} / 3.5 \mathrm{~hr} = 57.1 \mathrm{~km/hr}$.

Question 6

graph shows $\mathrm{v}_{\mathrm{x}}$ versus $\mathrm{t}$ for an object moving along straight line. What is the acceleration $a_{x}$ at $\mathrm{t}=11 \mathrm{~s}$ ?&#10; &#10;A) $-10 \mathrm{~m} / \mathrm{s}^{2}$&#10;B) $10 \mathrm{~m} / \mathrm{s}^{2}$&#10;C) $-22 \mathrm{~m} / \mathrm{s}^{2}$&#10;D) $22 \mathrm{~m} / \mathrm{s}^{2}$

Accepted Answer

The answer of graph shows $\mathrm{v}_{\mathrm{x}}$ versus $\mathrm{t}$ for an...

Question 7

figure shows the speedometer readings as a car comes to a stop. Solve graphically for the acceleration $a_{x}$ at $\mathrm{t}=7.0 \mathrm{~s}$.&#10; &#10;A) $2.0 \mathrm{~m} / \mathrm{s} 2$&#10;B) $-2.0 \mathrm{~m} / \mathrm{s}^{2}$&#10;C) $2.5 \mathrm{~m} / \mathrm{s}^{2}$&#10;D) $-2.5 \mathrm{~m} / \mathrm{s}^{2}$

Accepted Answer

The answer of figure shows the speedometer readings as a...

Question 8

figure shows the graph of $\mathrm{v}_{\mathrm{x}}$ versus time for an object moving along the $\mathrm{x}$-axis. What is the acceleration $a_{x}$ at $\mathrm{t}=7.0 \mathrm{~s}$ ?&#10; &#10;A) $0.4 \mathrm{~m} / \mathrm{s}^{2}$&#10;B) $0.5 \mathrm{~m} / \mathrm{s}^{2}$&#10;C) $4.0 \mathrm{~m} / \mathrm{s}^{2}$&#10;D) $5.0 \mathrm{~m} / \mathrm{s}^{2}$

Accepted Answer

The answer of figure shows the graph of $\mathrm{v}_{\mathrm{x}}$ versus...

Question 9

object starts from rest with an acceleration of $2.0 \mathrm{~m} / \mathrm{s}^{2}$ in the $+x$ -direction for 3.0 seconds. It then reduces its acceleration to $1.0 \mathrm{~m} / \mathrm{s}^{2}$ for 5.0 additional seconds. The total distance covered is about

A)

52 \mathrm{~m}

.
B)

11 \mathrm{~m}

.
C)

9.0 \mathrm{~m}

.
D)

7.6 \mathrm{~m}

.
E)

38 \mathrm{~m}

.

Accepted Answer

The total distance covered is calculated by adding the distances covered in each phase. For the first phase (3.0 seconds at $2.0 \mathrm{~m/s}^2$), the distance is $d_1 = \frac{1}{2}at^2 = \frac{1}{2} \times 2.0 \times (3.0)^2 = 9.0 \mathrm{~m}$. For the second phase (5.0 seconds at $1.0 \mathrm{~m/s}^2$), the distance is $d_2 = \frac{1}{2}at^2 = \frac{1}{2} \times 1.0 \times (5.0)^2 = 12.5 \mathrm{~m}$. However, the initial velocity for the second phase is the final velocity of the first phase, which is $v = at = 2.0 \times 3.0 = 6.0 \mathrm{~m/s}$. Thus, the correct calculation for the second phase should include the initial velocity: $d_2 = v_0t + \frac{1}{2}at^2 = 6.0 \times 5.0 + \frac{1}{2} \times 1.0 \times (5.0)^2 = 30.0 + 12.5 = 42.5 \mathrm{~m}$. Adding both distances: $9.0 \mathrm{~m} + 42.5 \mathrm{~m} = 51.5 \mathrm{~m}$, which is approximately $52 \mathrm{~m}$.

Question 10

$4.0 \mathrm{~kg}$ mass has a velocity of $12 \mathrm{~m} / \mathrm{s}$ to the WEST. The mass experiences a constant acceleration of 2.0 $\mathrm{m} / \mathrm{s}^{2}$ to the WEST for $3.0 \mathrm{sec}$ . What is the velocity of the mass at the end of the $3.0 \mathrm{sec}$ interval?

A)

18 \mathrm{~m} / \mathrm{s}

to the EAST
B)

0.0 \mathrm{~m} / \mathrm{s}

C)

18 \mathrm{~m} / \mathrm{s}

to the WEST
D)

6.0 \mathrm{~m} / \mathrm{s}

to the EAST
E)

6.0 \mathrm{~m} / \mathrm{s}

to the WEST

Accepted Answer

The final velocity can be calculated using the formula $v = u + at$, where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time. Substituting the given values: $v = 12 \, \text{m/s (to the WEST)} + (2 \, \text{m/s}^2 \times 3 \, \text{s}) = 12 \, \text{m/s} + 6 \, \text{m/s} = 18 \, \text{m/s}$ to the WEST.

Question 11

$4.0 \mathrm{~kg}$ mass has a velocity of $12 \mathrm{~m} / \mathrm{s}$ to the WEST. The mass experiences a constant acceleration of 2.0 $\mathrm{m} / \mathrm{s}^{2}$ to the WEST for $3.0 \mathrm{sec}$ . What is the velocity of the mass at the end of the $3.0 \mathrm{sec}$ interval?

A)

18 \mathrm{~m} / \mathrm{s}

to the EAST
B)

0.0 \mathrm{~m} / \mathrm{s}

C)

18 \mathrm{~m} / \mathrm{s}

to the WEST
D)

6.0 \mathrm{~m} / \mathrm{s}

to the EAST
E)

6.0 \mathrm{~m} / \mathrm{s}

to the WEST

Accepted Answer

The answer of $4.0 \mathrm{~kg}$ mass has a velocity of...

Question 12

car traveling at $3.0 \mathrm{~m} / \mathrm{s}$ has a constant acceleration of $4.0 \mathrm{~m} / \mathrm{s}^{2}$ in the same direction as the velocity. After 2.0 seconds, the speed is&#10;A) $5.0 \mathrm{~m} / \mathrm{s}$.&#10;B) $11 \mathrm{~m} / \mathrm{s}$.&#10;C) $7.0 \mathrm{~m} / \mathrm{s}$.&#10;D) $13 \mathrm{~m} / \mathrm{s}$.&#10;E) $9.0 \mathrm{~m} / \mathrm{s}$.

Accepted Answer

The answer of car traveling at $3.0 \mathrm{~m} / \mathrm{s}$...

Question 13

car traveling at $4.0 \mathrm{~m} / \mathrm{s}$ has a constant acceleration of $2.0 \mathrm{~m} / \mathrm{s}^{2}$ in the same direction as the velocity. After 3.0 seconds, the magnitude of the average velocity during the acceleration is

A)

5.0 \mathrm{~m} / \mathrm{s}

.
B)

9.0 \mathrm{~m} / \mathrm{s}

.
C)

7.0 \mathrm{~m} / \mathrm{s}

.
D)

11 \mathrm{~m} / \mathrm{s}

.
E) 13.

Accepted Answer

The answer of car traveling at $4.0 \mathrm{~m} / \mathrm{s}$...

Question 14

car traveling at $4.0 \mathrm{~m} / \mathrm{s}$ has a constant acceleration of $2.0 \mathrm{~m} / \mathrm{s}^{2}$ in the same direction as the velocity. After 3.0 seconds, the distance traveled is

A)

21 \mathrm{~m}

.
B)

13 \mathrm{~m}

.
C)

10 \mathrm{~m}

.
D)

17 \mathrm{~m}

.
E)

9 \mathrm{~m}

.

Accepted Answer

The answer of car traveling at $4.0 \mathrm{~m} / \mathrm{s}$...

Question 15

runner starts from rest and with an acceleration of $1.0 \mathrm{~m} / \mathrm{s}^{2}$ travels a distance of 10 meters. The time it takes the runner to cover the distance is&#10;A) $4.5 \mathrm{~s}$.&#10;B) $5.7 \mathrm{~s}$.&#10;C) $6.3 \mathrm{~s}$.&#10;D) $3.8 \mathrm{~s}$.&#10;E) $5.0 \mathrm{~s}$.

Accepted Answer

The answer of runner starts from rest and with an...

Question 16

runner starts from rest and with an acceleration of $2.0 \mathrm{~m} / \mathrm{s}^{2}$ travels a distance of 12 meters. The magnitude of the velocity of the runner at the end of the distance is&#10;A) $7.5 \mathrm{~m} / \mathrm{s}$.&#10;B) $3.4 \mathrm{~m} / \mathrm{s}$.&#10;C) $5.7 \mathrm{~m} / \mathrm{s}$.&#10;D) $6.9 \mathrm{~m} / \mathrm{s}$.&#10;E) $8.1 \mathrm{~m} / \mathrm{s}$.

Accepted Answer

The answer of runner starts from rest and with an...

Question 17

object starts with an initial velocity of magnitude $4.0 \mathrm{~m} / \mathrm{s}$ and accelerates at $4.0 \mathrm{~m} / \mathrm{s} 2$ in the same direction as the velocity for 6.0 seconds. It then accelerates at $2.0 \mathrm{~m} / \mathrm{s}^{2}$ in the opposite direction, until its velocity is the same as the initial value. What is the total distance covered? (Hint: make a graph of velocity versus time.)

A)

67 \mathrm{~m}

B)

216 \mathrm{~m}

C)

192 \mathrm{~m}

D)

96 \mathrm{~m}

E)

288 \mathrm{~m}

Accepted Answer

The answer of object starts with an initial velocity of...

Question 18

drive your car $5.0 \mathrm{~km}$ due east at $35 \mathrm{~km} / \mathrm{hr}$ and suddenly realize that you forgot your wallet. So, you return home, driving west at $40 \mathrm{~km} / \mathrm{hr}$ , and upon arrival you spend 10 minutes looking for it. Finally, you get back on the road and drive a total of $57.0 \mathrm{~km}$ due east. If your average velocity was $40 \mathrm{~km} / \mathrm{hr}$ for the whole journey, what was your average speed during the last leg?

A)

45 \mathrm{~km} / \mathrm{hr}

B)

40 \mathrm{~km} / \mathrm{hr}

C)

49 \mathrm{~km} / \mathrm{hr}

D)

58 \mathrm{~km} / \mathrm{hr}

Accepted Answer

The answer of drive your car $5.0 \mathrm{~km}$ due east...

Question 19

figure shows the graph of $v_{x}$ versus time for an object moving along the $x$-axis. Solve graphically for the distance traveled between $t=5.0 \mathrm{~s}$ and $\mathrm{t}=9.0 \mathrm{~s}$.&#10; &#10;A) $120 \mathrm{~m}$&#10;B) $130 \mathrm{~m}$&#10;C) $110 \mathrm{~m}$&#10;D) $100 \mathrm{~m}$

Accepted Answer

The answer of figure shows the graph of $v_{x}$ versus...

Question 20

figure shows the graph of $\mathrm{v}_{\mathrm{x}}$ versus time for an object moving along the $\mathrm{x}$-axis. Solve graphically for the average acceleration between $\mathrm{t}=5.0 \mathrm{~s}$ and $\mathrm{t}=9.0 \mathrm{~s}$.&#10; &#10;A) $4.0 \mathrm{~m} / \mathrm{s}^{2}$&#10;B) $5.0 \mathrm{~m} / \mathrm{s}^{2}$&#10;C) $0.4 \mathrm{~m} / \mathrm{s}^{2}$&#10;D) $0.5 \mathrm{~m} / \mathrm{s}^{2}$

Accepted Answer

The answer of figure shows the graph of $\mathrm{v}_{\mathrm{x}}$ versus...

Question 21

ball is thrown upward with a velocity of $19.6 \mathrm{~m} / \mathrm{s}$. What is its velocity after $3.00 \mathrm{~s}$ ?&#10;A) $9.80 \mathrm{~m} / \mathrm{s}$ down&#10;B) 19.6 down&#10;C) zero&#10;D) $9.80 \mathrm{~m} / \mathrm{s}$ up

Accepted Answer

The answer of ball is thrown upward with a velocity...

Question 22

ball is thrown straight up with an initial speed of $30 \mathrm{~m} / \mathrm{s}$. What is its speed after $4.2 \mathrm{~s}$ ?&#10;A) $42 \mathrm{~m} / \mathrm{s}$&#10;B) $72 \mathrm{~m} / \mathrm{s}$&#10;C) $30 \mathrm{~m} / \mathrm{s}$&#10;D) $11 \mathrm{~m} / \mathrm{s}$

Accepted Answer

The answer of ball is thrown straight up with an...

Question 23

ball is thrown straight up with a speed of $30.0 \mathrm{~m} / \mathrm{s}$. What is the maximum height reached by the ball?&#10;A) $132 \mathrm{~m}$&#10;B) $21.3 \mathrm{~m}$&#10;C) $92.0 \mathrm{~m}$&#10;D) $45.9 \mathrm{~m}$

Accepted Answer

The answer of ball is thrown straight up with a...

Question 24

sprinter runs $100.0 \mathrm{~m}$ in 12.2 seconds. If he travels at constant acceleration for the first $50.0 \mathrm{~m}$ and then at constant velocity for the final $50.0 \mathrm{~m}$ , what was his peak speed?

A)

8.20 \mathrm{~m} / \mathrm{s}

B)

17.5 \mathrm{~m} / \mathrm{s}

C)

4.10 \mathrm{~m} / \mathrm{s}

D)

12.3 \mathrm{~m} / \mathrm{s}

Accepted Answer

The answer of sprinter runs $100.0 \mathrm{~m}$ in 12.2 seconds....

Question 25

sprinter runs $100.0 \mathrm{~m}$ in 9.87 seconds. If he travels at constant acceleration for the first $75.0 \mathrm{~m}$ and then at constant velocity for the final $25.0 \mathrm{~m}$ , what was his acceleration during the first $75.0 \mathrm{~m}$ ?

A)

2.05 \mathrm{~m} / \mathrm{s}^{2}

B)

1.58 \mathrm{~m} / \mathrm{s}^{2}

C)

2.74 \mathrm{~m} / \mathrm{s}^{2}

D)

2.10 \mathrm{~m} / \mathrm{s}^{2}

Accepted Answer

The answer of sprinter runs $100.0 \mathrm{~m}$ in 9.87 seconds....

Question 26

electron from a heated filament in an electron gun travels with constant acceleration to the tip of the gun, where it is emitted. If the final speed of the electron is $1.5 \times 105 \mathrm{~m} / \mathrm{s}$ , the length of the gun from filament to tip is $1.25 \mathrm{~cm}$ , and the electron started from rest, what was the acceleration of the electron?

A)

9.0 \times 10^{9} \mathrm{~m} / \mathrm{s}^{2}

B)

1.8 \times 1012 \mathrm{~m} / \mathrm{s}^{2}

C)

9.0 \times 1011 \mathrm{~m} / \mathrm{s}^{2}

D)

6.0 \times 10^{6} \mathrm{~m} / \mathrm{s}^{2}

E)

9.0 \times 1010 \mathrm{~m} / \mathrm{s}^{2}

F)

1.8 \times 10^{9} \mathrm{~m} / \mathrm{s}^{2}

Accepted Answer

The answer of electron from a heated filament in an...

Question 27

ball is dropped at time $\mathrm{t}=0.0 \mathrm{~s}$ . At $\mathrm{t}=2.0 \mathrm{~s}$ , a second ball is thrown downward with speed $\mathrm{v}$ . What is $\mathrm{v}$ if at $\mathrm{t}=4.0 \mathrm{~s}$ , the two balls are at the same vertical position?

A)

25 \mathrm{~m} / \mathrm{s}

B)

29 \mathrm{~m} / \mathrm{s}

C)

49 \mathrm{~m} / \mathrm{s}

D)

15 \mathrm{~m} / \mathrm{s}

Accepted Answer

The answer of ball is dropped at time $\mathrm{t}=0.0 \mathrm{~s}$....

Deck 2: Motion Along a Line