Calculus 1 Quiz: Motion Position Velocity Acceleration

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Motion Position Velocity AccelerationQuestion 1 of 20

The position of a particle is given by $s(t) = t^3 - 6t^2 + 5$ for $t \ge 0$ . What is the maximum speed of the particle on the interval $1 \le t \le 4$ ?

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Calculus 1 Quiz

Calculus 1 Quiz: Motion Position Velocity Acceleration

Practice Motion Position Velocity Acceleration in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Motion Position Velocity Acceleration, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

The position of a particle is given by $s(t) = t^3 - 6t^2 + 5$ for $t \ge 0$ . What is the maximum speed of the particle on the interval $1 \le t \le 4$ ?

12 (correct answer)
0
9
16

Explanation: Speed is the absolute value of velocity,

|v(t)|

. First, find the velocity function:

v(t) = s'(t) = 3t^2 - 12t

. To find the maximum speed on

[1, 4]

, we must find the absolute maximum of

|v(t)|

. This will occur at an endpoint or where

v(t)

has a local extremum. Find the derivative of velocity:

a(t) = v'(t) = 6t - 12

. Set

a(t) = 0

to find critical points of

v(t)

6t-12=0 \implies t=2

. Now test the velocity at the endpoints and this critical point:

v(1) = 3-12 = -9

v(2) = 3(4)-12(2) = -12

v(4) = 3(16)-12(4) = 0

. The velocities on the interval range from -12 to 0. The corresponding speeds,

|v(t)|

, range from 0 to 12. The maximum speed is 12.

Question 2

The velocity of a particle is $v(t) = (t-1)(t-3)(t-5)$ for $t \ge 0$ . Which statement best describes the particle's motion at $t=2$ ?

Moving right and slowing down. (correct answer)
Moving right and speeding up.
Moving left and slowing down.
Moving left and speeding up.

Explanation: We need the signs of velocity

v(t)

and acceleration

a(t)

t=2

. At

t=2

v(2) = (2-1)(2-3)(2-5) = (1)(-1)(-3) = 3

. Since

v(2) > 0

, the particle is moving right. For acceleration, we first find the expression for

v(t) = t^3 - 9t^2 + 23t - 15

. Differentiating gives

a(t) = v'(t) = 3t^2 - 18t + 23

. Now evaluate at

t=2

a(2) = 3(2^2) - 18(2) + 23 = 12 - 36 + 23 = -1

. Since

v(2) > 0

and

a(2) < 0

, the signs are opposite, so the particle is slowing down. Thus, it is moving right and slowing down.

Question 3

The position of a particle on the x-axis is given by $s(t) = t^3 - 6t^2 + 9t + 1$ for $t \ge 0$ . What is the total distance traveled by the particle from $t=0$ to $t=2$ ?

6 (correct answer)
2
4
5

Explanation: Total distance requires accounting for any changes in direction. First, find the velocity:

v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)

. The particle changes direction when

v(t)

changes sign, which occurs at

t=1

and

t=3

. Within the interval

[0, 2]

, there is a change of direction at

t=1

. We must calculate the distance traveled on

[0, 1]

and

[1, 2]

separately. Positions are:

s(0) = 1

s(1) = 1-6+9+1=5

, and

s(2) = 8-24+18+1=3

. The distance traveled from

t=0

t=1

|s(1)-s(0)| = |5-1| = 4

. The distance traveled from

t=1

t=2

|s(2)-s(1)| = |3-5| = 2

. The total distance is

4+2=6

Question 4

The velocity of a particle moving along the x-axis is given by $v(t) = -t^3 + 6t^2 - 5$ for the time interval $0 \le t \le 5$ . At what time $t$ does the particle attain its maximum velocity?

$t=0$
$t=2$
$t=4$ (correct answer)
$t=5$

Explanation: To find the maximum velocity, we must find the absolute maximum of the function

v(t)

on the interval

[0, 5]

. We use the Extreme Value Theorem. First, find the derivative of

v(t)

, which is the acceleration

a(t)

a(t) = v'(t) = -3t^2 + 12t

. Set

a(t)=0

to find critical points:

-3t(t-4) = 0

, so

t=0

and

t=4

. Now, we test the endpoints and critical points in the velocity function:

v(0) = -5

v(4) = -(4^3)+6(4^2)-5 = -64+96-5 = 27

, and

v(5) = -(5^3)+6(5^2)-5 = -125+150-5=20

. The maximum value among these is 27, which occurs at

t=4

Question 5

A particle's position is described by $s(t) = |t^2 - 4|$ for $-3 \le t \le 3$ . Which statement is true about the particle's velocity at $t=2$ ?

The velocity is 0.
The velocity is 4.
The velocity is -4.
The velocity is undefined. (correct answer)

Explanation: The position function

s(t) = |t^2 - 4|

can be written as a piecewise function.

t^2-4

is non-negative for

t \le -2

t \ge 2

, and negative for

-2 < t < 2

. So,

s(t) = t^2-4

for

t \ge 2

and

s(t) = -(t^2-4) = 4-t^2

for

-2 < t < 2

. Velocity is the derivative of position,

v(t) = s'(t)

. To find the velocity at

t=2

, we must check if the derivative exists. The derivative from the right is

s'(t) = 2t

, so the right-hand limit of the derivative is

\lim_{t \to 2^+} 2t = 4

. The derivative from the left is

s'(t) = -2t

, so the left-hand limit of the derivative is

\lim_{t \to 2^-} -2t = -4

. Since the left-hand and right-hand derivatives are not equal, the derivative

s'(2)

does not exist. Therefore, the velocity is undefined at

t=2

Question 6

The position of a particle is $s(t) = 2\sin(t) - t$ for $0 \le t \le 2\pi$ . For which of the following values of $t$ is the particle moving to the right and slowing down?

$t = \pi/6$ (correct answer)
$t = \pi/3$
$t = 2\pi/3$
$t = \pi$

Explanation: Moving right means

v(t) > 0

. Slowing down means

v(t)

and

a(t)

have opposite signs. So we need

v(t) > 0

and

a(t) < 0

. First, find the functions:

v(t) = s'(t) = 2\cos(t) - 1

a(t) = v'(t) = -2\sin(t)

. Now we test the given points: A) At

t=\pi/6

v(\pi/6) = 2\cos(\pi/6) - 1 = 2(\sqrt{3}/2) - 1 = \sqrt{3}-1 > 0

a(\pi/6) = -2\sin(\pi/6) = -2(1/2) = -1 < 0

. Here,

v>0

and

a<0

, so the particle is moving right and slowing down. B) At

t=\pi/3

v=0

. C) At

t=2\pi/3

v<0

. D) At

t=\pi

v<0

Question 7

A particle moves on a line with velocity $v(t) = t^2 - kt + 4$ , where $k$ is a constant. The particle is momentarily at rest at $t=2$ . What is the acceleration of the particle at this instant?

-2
0 (correct answer)
2
4

Explanation: The particle is momentarily at rest when its velocity is zero. We are given this occurs at

t=2

, so

v(2)=0

. Substitute

t=2

into the velocity function:

v(2) = (2)^2 - k(2) + 4 = 0

. This gives

4 - 2k + 4 = 0

, so

8 - 2k = 0

, which means

k=4

. Now we know the full velocity function is

v(t) = t^2 - 4t + 4

. To find the acceleration, we differentiate the velocity:

a(t) = v'(t) = 2t - 4

. Finally, we evaluate the acceleration at

t=2

a(2) = 2(2) - 4 = 0

Question 8

A particle's velocity is given by $v(t) = \ln(t^2 - 4t + 5)$ for $t \ge 0$ . The particle is speeding up when:

$0 < t < 1$
$1 < t < 2$
$t > 2$ (correct answer)
$t > 3$

Explanation: The particle is speeding up when

v(t)

and

a(t)

have the same sign. First, analyze the sign of

v(t)

. The argument of the logarithm is

t^2-4t+5 = (t-2)^2 + 1

, which is always

\ge 1

. Since

\ln(x) > 0

for

x>1

v(t) > 0

for all

t \neq 2

. At

t=2

v(2) = \ln(1) = 0

. So, the particle is always moving right or is at rest. Now, find

a(t) = v'(t)

. Using the chain rule:

a(t) = \frac{1}{t^2-4t+5} \cdot (2t-4) = \frac{2(t-2)}{t^2-4t+5}

. Since the denominator is always positive, the sign of

a(t)

is determined by the sign of

2(t-2)

. Thus,

a(t) > 0

when

t>2

. Since

v(t)

is also positive for

t>2

, the particle is speeding up for

t>2

Question 9

The acceleration of a particle moving on a line is $a(t) = 12t^2 - 18t$ . At $t=0$ , the particle's velocity is $v(0)=5$ and its position is $s(0)=2$ . At $t=1$ , what is the particle's velocity?

-6
0 (correct answer)
1
5

Explanation: This question requires antidifferentiation. To find velocity

v(t)

, we integrate acceleration

a(t)

v(t) = \int (12t^2 - 18t) dt = 4t^3 - 9t^2 + C

. We use the initial condition

v(0)=5

to find C:

v(0) = 4(0)^3 - 9(0)^2 + C = 5 \implies C=5

. So the velocity function is

v(t) = 4t^3 - 9t^2 + 5

. Finally, we find the velocity at

t=1

v(1) = 4(1)^3 - 9(1)^2 + 5 = 4 - 9 + 5 = 0

. The information about position

s(0)

is extra information not needed to solve for velocity.

Question 10

Two particles, A and B, start at the origin and move along the x-axis. For $t \ge 0$ , their positions are given by $s_A(t) = t^3 - t$ and $s_B(t) = 5t^2 - 7t$ . At what time $t>0$ is the acceleration of particle A equal to the velocity of particle B?

$t=1$
$t=7/4$ (correct answer)
$t=5/3$
$t=2$

Explanation: First, find the required motion functions for each particle. For particle A, the velocity is

v_A(t) = s_A'(t) = 3t^2 - 1

and the acceleration is

a_A(t) = v_A'(t) = 6t

. For particle B, the velocity is

v_B(t) = s_B'(t) = 10t - 7

. The problem asks for the time

t

when the acceleration of A equals the velocity of B. So we set

a_A(t) = v_B(t)

6t = 10t - 7

. Solving for

t

gives

4t = 7

, so

t=7/4

Question 11

A particle moves along the x-axis with velocity $v(t) = 3(t-1)(t-3)$ for $t \ge 0$ . For which of the following time intervals is the particle's displacement equal to the total distance it has traveled?

$[0, 2]$
$[1, 3]$
$[2, 4]$
$[0, 1]$ (correct answer)

Explanation: A particle's displacement is equal to the total distance traveled over an interval if and only if the particle does not change direction on that interval. This means its velocity,

v(t)

, must be either always non-negative or always non-positive throughout the interval. The velocity function is

v(t) = 3(t-1)(t-3)

. The sign of

v(t)

changes at

t=1

and

t=3

v(t) \ge 0

for

t \in [0, 1] \cup [3, \infty)

v(t) \le 0

for

t \in [1, 3]

. Let's check the given intervals:

On $[0, 1]$ , $v(t) \ge 0$ , so the particle only moves in the positive direction. Displacement equals total distance.
On $[1, 3]$ , $v(t) \le 0$ , so the particle only moves in the negative direction. Displacement is negative while total distance is positive, so they are not equal.
On $[0, 2]$ , the particle moves in the positive direction for $[0, 1]$ and in the negative direction for $[1, 2]$ . Since it changes direction, displacement is not equal to total distance.
On $[2, 4]$ , the particle moves in the negative direction for $[2, 3]$ and in the positive direction for $[3, 4]$ . Since it changes direction, displacement is not equal to total distance.

Question 12

For a particle in rectilinear motion, its velocity $v(t)$ is a differentiable function of time $t$ . For a specific time interval $(a, b)$ , it is known that $v(t) < 0$ and the acceleration $a(t) > 0$ . Which of the following statements must be true for the particle on the interval $(a, b)$ ?

The particle's speed is increasing.
The particle's speed is decreasing. (correct answer)
The particle's position from the origin is increasing.
The particle changes its direction of motion.

Explanation: The speed of the particle is given by

|v(t)|

. A particle is speeding up if its velocity and acceleration have the same sign, and it is slowing down if they have opposite signs.

We are given that $v(t) < 0$ , which means the particle is moving in the negative direction. Consequently, its position is decreasing, not increasing.
We are given that $a(t) > 0$ .
Since $v(t)$ and $a(t)$ have opposite signs, the particle is slowing down. Its speed, $|v(t)|$ , is decreasing.
For the particle to change direction, its velocity $v(t)$ must change sign. We are given that $v(t) < 0$ for the entire interval $(a, b)$ , so it does not change direction.

Question 13

A particle moves along a line such that its position is given by $s(t) = t^3 - 6t^2 + 9t + 1$ for $t \ge 0$ . The particle is momentarily at rest at two different times. What is the magnitude of the difference in its acceleration at these two times?

$0$
$6$
$12$ (correct answer)
$2$

Explanation: First, find the velocity function

v(t) = s'(t)

v(t) = 3t^2 - 12t + 9

. The particle is momentarily at rest when

v(t) = 0

3t^2 - 12t + 9 = 0

t^2 - 4t + 3 = 0

(t-1)(t-3) = 0

The particle is at rest at

t=1

and

t=3

. Next, find the acceleration function

a(t) = v'(t)

a(t) = 6t - 12

. Now, calculate the acceleration at the two times when the particle is at rest:

a(1) = 6(1) - 12 = -6

a(3) = 6(3) - 12 = 18 - 12 = 6

. The difference in the accelerations is

a(3) - a(1) = 6 - (-6) = 12

. The magnitude of this difference is

|12| = 12

Question 14

The position of a particle is given by $s(t) = t^3 - 12t^2 + 36t$ for $t \ge 0$ . At what time $t$ in the open interval $(0, 3)$ is the particle's instantaneous velocity equal to its average velocity on the closed interval $[0, 3]$ ?

$4 - \sqrt{7}$ (correct answer)
$4 + \sqrt{7}$
$2$
$4 - \sqrt{5}$

Explanation: This question is an application of the Mean Value Theorem. First, calculate the average velocity on the interval

[0, 3]

s(0) = 0^3 - 12(0)^2 + 36(0) = 0

s(3) = 3^3 - 12(3)^2 + 36(3) = 27 - 108 + 108 = 27

. The average velocity is

v_{avg} = \frac{s(3) - s(0)}{3 - 0} = \frac{27}{3} = 9

. Next, find the instantaneous velocity function,

v(t) = s'(t)

v(t) = 3t^2 - 24t + 36

. Set the instantaneous velocity equal to the average velocity:

3t^2 - 24t + 36 = 9

3t^2 - 24t + 27 = 0

t^2 - 8t + 9 = 0

Use the quadratic formula to solve for

t

t = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(9)}}{2(1)} = \frac{8 \pm \sqrt{64 - 36}}{2} = \frac{8 \pm \sqrt{28}}{2} = \frac{8 \pm 2\sqrt{7}}{2} = 4 \pm \sqrt{7}

. The two possible times are

t = 4 - \sqrt{7}

and

t = 4 + \sqrt{7}

. We must determine which of these lies in the interval

(0, 3)

. Since

2 < \sqrt{7} < 3

, we have:

t_1 = 4 - \sqrt{7} \approx 4 - 2.65 = 1.35

. This is in

(0, 3)

t_2 = 4 + \sqrt{7} \approx 4 + 2.65 = 6.65

. This is not in

(0, 3)

. Therefore, the correct time is

t = 4 - \sqrt{7}

Question 15

A particle moves along a line with position $s(t) = t^3 - kt^2 + 8$ , where $k$ is a positive constant. The particle changes direction at $t=2$ . What is the acceleration of the particle at $t=k$ ?

$0$
$6$
$12$ (correct answer)
$18$

Explanation: First, find the velocity function:

v(t) = s'(t) = 3t^2 - 2kt

. A particle changes direction when its velocity changes sign. This can only happen when

v(t)=0

. We are given that this occurs at

t=2

. So, we set

v(2)=0

and solve for

k

v(2) = 3(2^2) - 2k(2) = 12 - 4k = 0

. Solving for

k

gives

4k = 12

, so

k=3

. Now we have the specific position function

s(t) = t^3 - 3t^2 + 8

. We need to find the acceleration at

t=k=3

. First, find the acceleration function:

a(t) = v'(t) = 6t - 2k

. Since we found

k=3

, the acceleration function is

a(t) = 6t - 6

. Now, evaluate the acceleration at

t=k=3

a(3) = 6(3) - 6 = 18 - 6 = 12

Question 16

A particle's position is given by $s(t) = 2t^3 - 21t^2 + 60t$ for $t \ge 0$ . What is the particle's velocity when its acceleration is zero?

-25.5
-13.5 (correct answer)
3.5
60

Explanation: This is a two-step problem. First, find the time when acceleration is zero. Then, find the velocity at that time.

s(t) = 2t^3 - 21t^2 + 60t

v(t) = s'(t) = 6t^2 - 42t + 60

a(t) = v'(t) = 12t - 42

. Set

a(t) = 0

12t - 42 = 0 \implies t = 42/12 = 7/2 = 3.5

. Now, substitute this time back into the velocity function:

v(3.5) = 6(3.5)^2 - 42(3.5) + 60 = 6(12.25) - 147 + 60 = 73.5 - 147 + 60 = -13.5

Question 17

Particle P has position $x_P(t) = \sin(\pi t)$ and Particle Q has position $x_Q(t) = t^2 - t$ . For the time interval $0 < t < 2$ , on which interval(s) are the particles moving in opposite directions?

$(0, 1/2) \cup (3/2, 2)$
$(1/2, 3/2)$ only
$(0, 1/2) \cup (1/2, 3/2)$ (correct answer)
$(1/2, 2)$

Explanation: Particles move in opposite directions when their velocities have opposite signs. First, find the velocities:

v_P(t) = \pi \cos(\pi t)

and

v_Q(t) = 2t - 1

. We need to analyze the signs of each. For

v_P(t)

, the sign changes when

\cos(\pi t) = 0

, which is at

t=1/2

and

t=3/2

. For

v_Q(t)

, the sign changes when

2t-1=0

, which is at

t=1/2

. Let's test the intervals: (1) For

t \in (0, 1/2)

v_P(t) > 0

and

v_Q(t) < 0

. Opposite signs. (2) For

t \in (1/2, 3/2)

v_P(t) < 0

and

v_Q(t) > 0

. Opposite signs. (3) For

t \in (3/2, 2)

v_P(t) > 0

and

v_Q(t) > 0

. Same signs. Therefore, they move in opposite directions on

(0, 1/2) \cup (1/2, 3/2)

Question 18

A particle moves along the x-axis with velocity $v(t) = e^{-t} (2t - t^2)$ for $t > 0$ . The particle's acceleration is zero at the exact moment its velocity reaches a local maximum. At what time $t$ does this occur?

$t = 2 - \sqrt{2}$
$t = 2$
$t = 2 + \sqrt{2}$ (correct answer)
$t = 1$

Explanation: The particle's acceleration is zero when its velocity reaches a local maximum. We need to find the time

t

when

a(t) = v'(t) = 0

. Using the product rule for

v(t) = e^{-t} (2t - t^2)

a(t) = (e^{-t})' (2t - t^2) + e^{-t} (2t - t^2)' = -e^{-t}(2t - t^2) + e^{-t}(2 - 2t)

. Set

a(t) = 0

e^{-t} [-(2t - t^2) + (2 - 2t)] = 0

. Since

e^{-t} > 0

, we solve

-2t + t^2 + 2 - 2t = 0 \implies t^2 - 4t + 2 = 0

. Using the quadratic formula:

t = \frac{4 \pm \sqrt{16 - 4(1)(2)}}{2} = \frac{4 \pm \sqrt{8}}{2} = 2 \pm \sqrt{2}

. To determine which is a maximum, we can use the second derivative test for velocity,

a'(t)

, or analyze the sign of

a(t)

. The parabola

y=t^2-4t+2

opens up, so it changes from + to - to + at its roots. The acceleration

a(t)

goes from positive to negative at the larger root,

t = 2 + \sqrt{2}

, which indicates a local maximum for velocity.

Question 19

A particle moves along a line with acceleration $a(t) = 2t$ . Its velocity at $t=3$ is $v(3)=5$ . What is its average velocity over the interval $[0, 3]$ ?

-4
-1 (correct answer)
5/3
3

Explanation: First, find the velocity function by integrating acceleration:

v(t) = \int 2t dt = t^2 + C

. Use the condition

v(3)=5

to find C:

3^2+C=5 \implies 9+C=5 \implies C=-4

. So,

v(t) = t^2-4

. Average velocity over

[0, 3]

is given by

\frac{s(3)-s(0)}{3-0}

, which requires finding the position function

s(t)

s(t) = \int (t^2-4) dt = \frac{1}{3}t^3 - 4t + D

. The displacement is

s(3)-s(0) = (\frac{1}{3}(3)^3 - 4(3) + D) - (\frac{1}{3}(0)^3 - 4(0) + D) = (9-12) - 0 = -3

. The average velocity is

\frac{-3}{3-0} = -1

Question 20

A particle moves along a straight line. For a certain time interval, the particle's velocity is decreasing, but its speed is increasing. Which of the following statements must be true during that interval?

The particle's velocity is positive and its acceleration is negative.
The particle's velocity is negative and its acceleration is positive.
The particle's velocity is positive and its acceleration is positive.
The particle's velocity is negative and its acceleration is negative. (correct answer)

Explanation: 'Velocity is decreasing' means that the derivative of velocity, acceleration, is negative (

a(t) < 0

). 'Speed is increasing' means that velocity and acceleration have the same sign. Since we know acceleration is negative, velocity must also be negative (

v(t) < 0

). Therefore, the particle's velocity is negative and its acceleration is negative. This corresponds to moving in the negative direction and speeding up.