Question 1

An object starts from rest and travels in a straight line with an acceleration of 2.0 m/s² for 3.0 seconds. It then reduces its acceleration to 1.0 m/s² for 5.0 additional seconds. The magnitude of the velocity at the end of the 5.0 second interval is

Accepted Answer

To find the magnitude of the velocity at the end of the 8.0-second interval (3.0 seconds at 2.0 m/s^2 and 5.0 seconds at 1.0 m/s^2), we calculate the velocity at the end of each interval and sum them. For the first interval: $v = u + at = 0 + (2.0 \, m/s^2)(3.0 \, s) = 6.0 \, m/s$. For the second interval, the initial velocity is 6.0 m/s, so $v = u + at = 6.0 \, m/s + (1.0 \, m/s^2)(5.0 \, s) = 6.0 \, m/s + 5.0 \, m/s = 11.0 \, m/s$.

Question 2

The figure shows the graph of v_x versus time for an object moving along the x-axis. What is the acceleration a_x at t = 7.0 s?

Accepted Answer

The answer of The figure shows the graph of v_x...

Question 3

A car traveling at 4.0 m/s has a constant acceleration of 2.0 m/s² in the same direction as the velocity. After 3.0 seconds, the magnitude of the average velocity during the acceleration is

Accepted Answer

The final velocity after 3 seconds of constant acceleration of 2.0 m/s² is:
$v_f = v_i + at = 4.0 	ext{ m/s} + 2.0 	ext{ m/s}² \cdot 3.0 	ext{ s} = 10.0 	ext{ m/s}$
The average velocity during this time can be found by taking the average of the initial and final velocities:
$v_{avg} = \frac{v_i + v_f}{2} = \frac{4.0 	ext{ m/s} + 10.0 	ext{ m/s}}{2} = 7.0 	ext{ m/s}$
Therefore, the answer is B.

Question 4

A runner starts from rest and with an acceleration of 1.0 m/s² travels a distance of 10 meters. The time it takes the runner to cover the distance is

Accepted Answer

Using the equation of motion s = ut + 1/2 at^2, where s is the distance, u is the initial velocity (which is 0 m/s), a is the acceleration, and t is the time:

s = ut + 1/2 at^2
10 = 0 + 1/2 (1.0) t^2
20 = t^2
t = 4.5 seconds

Therefore, the time it takes the runner to cover 10 meters is 4.5 seconds. Option D is the correct answer.

Question 5

A runner starts from rest and with an acceleration of 2.0 m/s² travels a distance of 12 meters. The magnitude of the velocity of the runner at the end of the distance is

Accepted Answer

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

Question 6

A 4.0 kg mass has a velocity of 10 m/s to the EAST. The mass undergoes a constant acceleration of 4.0 m/s² to the WEST for 3.0 sec. What is the velocity of the mass at the end of the 3.0 sec interval?

Accepted Answer

Using the equation v = u + at, where u is initial velocity, a is acceleration, t is time, and v is final velocity, we can solve for the final velocity of the mass after 3.0 sec:
v = 10 m/s (initial velocity) - 4.0 m/s^2 (acceleration) x 3.0 sec
v = 10 m/s - 12 m/s
v = -2 m/s to the WEST
Therefore, the answer is B) 2.0 m/s to the WEST.

Question 7

A boat is traveling at 4.0 m/s as it passes the starting line of a race. If the boat accelerates at 1.0 m/s² in the same direction as the velocity for 6.0 seconds, then the distance the boat has traveled after 6.0 seconds is

Accepted Answer

The distance traveled by the boat can be calculated using the formula for distance under constant acceleration: $d = v_0t + \frac{1}{2}at^2$, where $v_0$ is the initial velocity (4.0 m/s), $a$ is the acceleration (1.0 m/s$^2$), and $t$ is the time (6.0 seconds). Plugging in these values gives $d = 4.0 \times 6.0 + \frac{1}{2} \times 1.0 \times 6.0^2 = 24 + 18 = 42$ meters.

Question 8

The graph shows v_xversus t for an object moving along straight line. What is the acceleration a_x at t = 11 s?

Accepted Answer

The answer of The graph shows v_xversus t for...

Question 9

A boat is traveling at 4.0 m/s as it passes the starting line of a race. If the boat accelerates at 2.0 m/s² for 3.0 seconds, in the same direction as the velocity, then the magnitude of the velocity of the boat after the 3.0 seconds is

Accepted Answer

To solve this problem, we can use the equation:

v = v0 + at

where:

- v is the final velocity
- v0 is the initial velocity
- a is the acceleration
- t is the time interval

In this problem, we know that:

- v0 = 4.0 m/s
- a = 2.0 m/s^2
- t = 3.0 s

Therefore:

v = 4.0 m/s + 2.0 m/s^2 x 3.0 s = 10 m/s

So the magnitude of the velocity of the boat after 3.0 seconds is 10 m/s. Therefore, the correct answer is (D).

Question 10

A car traveling at 4.0 m/s has a constant acceleration of 2.0 m/s² in the same direction as the velocity. After 3.0 seconds, the distance traveled is

Accepted Answer

The distance traveled by the car can be calculated using the formula for distance under constant acceleration: $d = v_0t + \frac{1}{2}at^2$, where $v_0$ is the initial velocity, $a$ is the acceleration, and $t$ is the time. Substituting the given values: $d = 4.0 \, \text{m/s} \times 3.0 \, \text{s} + \frac{1}{2} \times 2.0 \, \text{m/s}^2 \times (3.0 \, \text{s})^2 = 12 \, \text{m} + 9 \, \text{m} = 21 \, \text{m}$.

Question 11

The graph shows v_xversus t for an object moving along straight line. What is the average speed from t = 0 to t = 11 s?

Accepted Answer

The answer of The graph shows v_xversus t for...

Question 12

A car starts from rest and travels a distance of 100 m in 15 seconds with a constant acceleration. The magnitude of the average velocity of the car for the 15 second interval is

Accepted Answer

The answer of A car starts from rest and travels...

Question 13

A car starts from rest and travels a distance of 100 m in 20 seconds, with constant acceleration. The magnitude of the velocity of the car is at the end of the 20 second interval is

Accepted Answer

The velocity can be found using the formula for constant acceleration: $v = u + at$, where $v$ is the final velocity, $u$ is the initial velocity (0 m/s since the car starts from rest), $a$ is the acceleration, and $t$ is the time. To find the acceleration, we use the formula $s = ut + \frac{1}{2}at^2$, where $s$ is the distance (100 m), $u$ is the initial velocity (0 m/s), and $t$ is the time (20 s). Substituting the given values, we get $100 = 0 + \frac{1}{2}a(20)^2$, solving for $a$ gives $a = 0.5$ m/s$^2$. Substituting $a$ back into the first formula gives $v = 0 + 0.5 \times 20 = 10$ m/s.

Question 14

The graph shows v_x versus t for an object moving in a straight line. What is the average speed from t = 0 s to t = 9 s?

Accepted Answer

The answer of The graph shows v_x versus t for...

Question 15

A car traveling at 3.0 m/s has a constant acceleration of 4.0 m/s² in the same direction as the velocity. After 2.0 seconds, the speed is

Accepted Answer

The answer of A car traveling at 3.0 m/s has...

Question 16

A car starts from rest and travels a distance of 100 m in 10 seconds, with constant acceleration. The magnitude of the acceleration of the car is

Accepted Answer

The formula to calculate acceleration when a car starts from rest and travels a certain distance with constant acceleration is $a = \frac{2s}{t^2}$, where $s$ is the distance traveled and $t$ is the time taken. Substituting $s = 100$ m and $t = 10$ s, we get $a = \frac{2 \times 100}{10^2} = \frac{200}{100} = 2.0$ m/s$^2$.

Question 17

A boat is traveling at 4.0 m/s as it passes the starting line of a race. If the boat accelerates at 1.0 m/s² for 6.0 seconds, in the same direction as the velocity, then the magnitude of the average velocity of the boat during the 6.0 seconds is

Accepted Answer

The answer of A boat is traveling at 4.0 m/s...

Question 18

The figure shows the speedometer readings as a car comes to a stop. Solve graphically for the acceleration a_x at t = 7.0 s.

Accepted Answer

To solve for acceleration, we need to use the equation: 
a = (v_f - v_i) / t

At t = 7.0 s, the speed of the car is approximately 15 m/s. We can find the initial speed by looking at the graph, which is approximately 39 m/s.

Plugging in the values, we get: 
a = (15 m/s - 39 m/s) / 6 s = -4 m/s^2

However, this is the acceleration of the car, not the acceleration of the speedometer needle. To find the acceleration of the speedometer needle, we need to look at the slope of the tangent to the graph at t = 7.0 s.

Drawing a tangent line to the curve at t = 7.0 s, we can see that it has a slope of approximately -2.5 m/s^2. This is the acceleration of the speedometer needle, but it is in the negative direction because the speedometer is decreasing.

Therefore, the answer is B) -2.5 m/s^2.

Question 19

An object starts from rest with an acceleration of 2.0 m/s² in the +x-direction for 3.0 seconds. It then reduces its acceleration to 1.0 m/s² for 5.0 additional seconds. The total distance covered is about

Accepted Answer

The total distance covered can be calculated by finding the distance covered in each phase of the motion and then summing these distances. For the first phase (first 3.0 seconds), the object starts from rest and accelerates at 2.0 m/s^2. The distance covered in this phase can be calculated using the formula $d_1 = \frac{1}{2}at^2$, where $a = 2.0 \, \text{m/s}^2$ and $t = 3.0 \, \text{s}$, giving $d_1 = \frac{1}{2} \times 2.0 \times (3.0)^2 = 9.0 \, \text{m}$. For the second phase (next 5.0 seconds), the acceleration is 1.0 m/s^2. Using the same formula, $d_2 = \frac{1}{2} \times 1.0 \times (5.0)^2 = 12.5 \, \text{m}$. However, the initial velocity for the second phase is not zero; it's the final velocity from the first phase, which is $v = at = 2.0 \times 3.0 = 6.0 \, \text{m/s}$. The distance covered due to this initial velocity over 5.0 seconds is $v \times t = 6.0 \times 5.0 = 30.0 \, \text{m}$. Thus, the total distance for the second phase is $d_2 + 30.0 = 12.5 + 30.0 = 42.5 \, \text{m}$. Summing the distances from both phases, $d_1 + d_2 = 9.0 + 42.5 = 51.5 \, \text{m}$, which is approximately 52 m.

Question 20

A 4.0 kg mass has a velocity of 12 m/s to the WEST. The mass experiences a constant acceleration of 2.0 m/s² to the WEST for 3.0 sec. What is the velocity of the mass at the end of the 3.0 sec interval?

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: $v = 12\,m/s + (2\,m/s^2 \times 3\,s) = 12\,m/s + 6\,m/s = 18\,m/s$ to the WEST.

Quiz 2: Motion Along a Line

An object starts from rest and travels in a straight line with an acceleration of 2.0 m/s2 for 3.0 seconds. It then reduces its acceleration to 1.0 m/s2 for 5.0 additional seconds. The magnitude of the velocity at the end of the 5.0 second interval is

The figure shows the graph of vx versus time for an object moving along the x-axis. What is the acceleration ax at t = 7.0 s?

A car traveling at 4.0 m/s has a constant acceleration of 2.0 m/s2 in the same direction as the velocity. After 3.0 seconds, the magnitude of the average velocity during the acceleration is

A runner starts from rest and with an acceleration of 1.0 m/s2 travels a distance of 10 meters. The time it takes the runner to cover the distance is

A runner starts from rest and with an acceleration of 2.0 m/s2 travels a distance of 12 meters. The magnitude of the velocity of the runner at the end of the distance is

A 4.0 kg mass has a velocity of 10 m/s to the EAST. The mass undergoes a constant acceleration of 4.0 m/s2 to the WEST for 3.0 sec. What is the velocity of the mass at the end of the 3.0 sec interval?

A boat is traveling at 4.0 m/s as it passes the starting line of a race. If the boat accelerates at 1.0 m/s2 in the same direction as the velocity for 6.0 seconds, then the distance the boat has traveled after 6.0 seconds is

The graph shows vx versus t for an object moving along straight line. What is the acceleration ax at t = 11 s?

A boat is traveling at 4.0 m/s as it passes the starting line of a race. If the boat accelerates at 2.0 m/s2 for 3.0 seconds, in the same direction as the velocity, then the magnitude of the velocity of the boat after the 3.0 seconds is

A car traveling at 4.0 m/s has a constant acceleration of 2.0 m/s2 in the same direction as the velocity. After 3.0 seconds, the distance traveled is

The graph shows vx versus t for an object moving along straight line. What is the average speed from t = 0 to t = 11 s?