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Resources

Relate position, velocity, speed, and acceleration using derivatives

physics C textbook

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ext Table Problems (Derivatives)

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1

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The velocity of an object moving in a straight line, in kilometers fer hour, can be modeled by the function

v(t)

t

is measured in hours. Selected values of

v(t)

are shown in the table below. Use the data in the table to estimate the acceleration of the object at time

t=4

, using proper units.

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\begin{tabular}{|c|c|c|c|c|c|}

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t

0

2

6

12

17

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v(t)

4

9

2

10

5

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1500 \mathrm{Nm}

acting on a body produces an angular acceleration of

3.2 \mathrm{rad} / \mathrm{s}^{2}

. Find M.I. of the body. [Mar

09

Below is the graph of a trigonometric function. It intersects its midline at

(\pi\tau, 2.75)

, and it has a maximum point at

(\frac{5}{4}\pi\tau, 6.75)

what is the amplitude of the function

Below is the graph of a trigonometric function. It intersects its midline at

(\pi\tau, 2.75)

, and it has a maximum point at

(\frac{5}{4}\pi\tau, 6.75)

what is the amplitude of the function

1

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The graph shows velocity

v

as a function of time

t

4

seconds of a carts motion on a horizontal track. Which of the following best describes the acceleration of the cart during the time interval shown?

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(A) The acceleration is zero during the entire intervat:

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(B) The acceleration is nonzero and constant during the entire interval.

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CC) The acceleration is initially nonzero and constant and then is zero.

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(D) The acceleration is initially increasing and then is zeto.

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(E) The acceleration is intially increasing and then is nonzero and constant.

An object travels along a straight line. The function

v(t) = 5\cos t - 4t

gives the object's velocity, in feet per hour, at time

t

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Write a function that gives the object's acceleration

a(t)

in feet per hour per hour.

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An object travels along a straight line. The function

v(t) = 2\sqrt{t} - 6\sqrt[3]{t}

gives the object's velocity, in meters per second, at time

t

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Write a function that gives the object's acceleration

a(t)

in meters per second per second.

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An object travels along a straight line. The function

s(t) = 3t^{\frac{4}{3}} + 6t^{\frac{1}{3}}

gives the object's position, in feet, at time

t

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Write a function that gives the object's velocity

v(t)

in feet per minute.

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v(t) =

An object travels along a straight line. The function

v(t) = -13t^3 - 2t^2 + 8t + 9

gives the object's velocity, in feet per second, at time

t

seconds. Write a function that gives the object's acceleration

a(t)

in feet per second per second.

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An object travels along a straight line. The function

s(t) = -32t^2 - 6t + 45

gives the object's position, in feet, at time

t

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Write a function that gives the object's velocity

v(t)

in feet per second.

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An object travels along a straight line. The function

s(t) = 10t - 3

gives the object's position, in kilometers, at time

t > 0

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Write a function that gives the object's velocity

v(t)

in kilometers per hour.

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v(t) = \underline{\hspace{3cm}}

An object travels along a straight line. The function

v(t) = - t^{\frac{1}{5}} + 2t^{-2}

gives the object's velocity, in feet per minute, at time

t > 0

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Write a function that gives the object's acceleration

a(t)

in feet per minute per minute.

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An object travels along a straight line. The function

s(t) = -9.8t^2 + 18t + 100

gives the object's position, in feet, at time

t

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Write a function that gives the object's velocity

v(t)

in feet per second.

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An object travels along a straight line. The function

v(t) = 4\sqrt{t} - 3

gives the object's velocity, in kilometers per hour, at time

t > 0

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Write a function that gives the object's acceleration

a(t)

in kilometers per hour per hour.

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An object travels along a straight line. The function

v(t) = 4t^4 - 3t^3 - 11t + 3

gives the object's velocity, in feet per hour, at time

t

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Write a function that gives the object's acceleration

a(t)

in feet per hour per hour.

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An object travels along a straight line. The function

v(t) = 3\sqrt{t} - 4t^2 + t

gives the object's velocity, in feet per hour, at time

t

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Write a function that gives the object's acceleration

a(t)

in feet per hour per hour.

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An object travels along a straight line. The function

s(t) = t^2 - 10t + 12

gives the object's position, in miles, at time

t

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Write a function that gives the object's velocity

v(t)

in miles per hour.

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An object travels along a straight line. The function

s(t) = 5t^3 + 12t^2 - 15t - 10

gives the object's position, in feet, at time

t

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Write a function that gives the object's velocity

v(t)

in feet per second.

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v(t) =

An object travels along a straight line. The function

v(t) = 4t^3 - 12t^2 + 5t + 10

gives the object's velocity, in miles per hour, at time

t

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Write a function that gives the object's acceleration

a(t)

in miles per hour per hour.

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An object travels along a straight line. The function

s(t) = 3\cos t - 7e^t

gives the object's position, in feet, at time

t

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Write a function that gives the object's velocity

v(t)

in feet per second.

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An object travels along a straight line. The function

s(t) = -4t^5 + 10t^4 + 7t^3 + t + 11

gives the object's position, in meters, at time

t

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Write a function that gives the object's velocity

v(t)

in meters per second.

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v(t) =

Madison started hiking at an elevation of

\mathbf{4}

meters above sea level. She hiked up at a rate of

\mathbf{3}

meters per minute. The graph below represents the relationship between the number of minutes and the elevation.

4

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What is the slope of the line?

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In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is

0.170\,\text{s}

. The frequency of the wave is

The potential energy of a particle varies the distance

x

from a fixed origin as

U= \frac{A}{x^2} +B

A

B

are dimensional constants, then find the dimensional formula for

\frac{A^2}{B^2}.

If the velocity of a particle moving along the

x

v(t)=2 t-4

t=0

its position is

4

, then at any time

t

x(t)

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t^{2}-4 t

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t^{2}-4 t-4

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t^{2}-4 t+4

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2 t^{2}-4 t

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2 t^{2}-4 t+4

1

. A jeeprey acceleration of a speed of

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

over a distance of

300 \mathrm{~m}

. Determine the aceleration (assume uniformly) af the jeeprey.

3

. The distance a person can see through a particular submarine periscope is given by the equation

d=6 \sqrt{h-4}+3

h

is the height in feet above water.

A cheetah accelerating uniformly from rest reaches a speed of

29\,\text{ms}^{-1}

2.0\,\text{s}

and then maintains this speed for

15\,\text{seconds}

. (a) Calculate (i) its acceleration (ii) the distance it travels while accelerating

3

. A block with a mass of

30 \mathrm{~kg}

is located on a horizontal, frictionless tabletop. This block is connected by a rope to another block with a mass of

10 \mathrm{~kg}

. The rope is looped through a pulley on the table's edge so that the less massive block is hanging over the edge. What is the magnitude of the acceleration of the larger block across the table?

14

. The equation for wind speed

w

, in miles per hour, near the center of a tornado is

w=93 \log _{10}(d)+65

d

is the distance in miles that the tornado travels.

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A. Write this equation in exponential form.

P

moves in a straight line with displacement relative to origin given by

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s=2 \sin (\pi t)+\sin (2 \pi t), t \geq 0,

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t

is the time in seconds and the displacement is measured in centimetres.

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(i) Write down the period of the function

s

A particle moves along the

x

-axis with velocity

v(t)=\frac{t^{2}}{\sin ^{2}(t)+2}

. The particle is at position

x=3

t=2

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What is the particle's position at time

t=7

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Use a graphing calculator and round your answer to three decimal places.

A particle moves along the

x

-axis with velocity

v(t)=\cos \left(t^{2}-3 t\right)

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What is the total distance the particle has traveled between the times

t=3

t=6

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Use a graphing calculator and round your answer to three decimal places.

A particle moves along the

x

-axis with velocity

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v(t)=\ln \left(t^{2}+5 t+1\right) \text {. }

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What is the particle's displacement between the times

t=1

t=5

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Use a graphing calculator and round your answer to three decimal places.

Herman is riding his hoverboard. The function

V

gives Herman's velocity (in meters per second),

t

seconds after he started riding.

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What is the best interpretation for the following statement?

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The value of the derivative of

V

t=5

1

5

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1

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A) During the first

5

seconds, Herman's acceleration is

1

5

meters per second squared.

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5

seconds, Herman's acceleration is

1

5

meters per second squared.

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5

seconds, Herman's velocity is

1

5

meters per second.

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5

seconds, Herman's acceleration is

1

5

x=v_{\circ} t+\frac{1}{2} a t^{2}

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The horizontal displacement,

x

, of an object with constant acceleration,

a

, initial velocity,

v_{\circ}

, at elapsed time,

t

, is given by the equation. Which of the following equations correctly shows the acceleration in terms of displacement, initial velocity, and time?

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1

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a=\frac{2 \sqrt{x-v_{\circ} t}}{t}

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a=\frac{2\left(x-v_{0}\right)}{t}

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a=\frac{2\left(x-v_{\circ} t\right)}{t^{2}}

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a=\frac{2 x}{v_{0} t^{3}}

A big ship drops its anchor.

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E

represents the anchor's elevation relative to the water's surface (in meters) as a function of time

t

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E=-2.4 t+75

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How far does the anchor drop every

5

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A tug is moxing hack and forth on a straight path. The velocity of the bug is given by

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v(t)=t^{2}-3t

. Find the average acceleration of the bug on the interval

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[1,4]

A hug is mowing hack and forth on a straight path. The velocity of the bug is given by

v(t)=t^{2}-3 t

. Find the average acceleration of the bug on the interval

[1,4]

A bug is moving back and forth on a straight path. The velocity of the bug is given by

v(t)=t^{2}-3 t

. Find the average acceleration of the bug on the interval

[1,4]

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\left|\frac{4^{3}}{3}-24\right|\left|\frac{t^{3}}{3}-\frac{3 t^{2}}{2}\right|_{1}^{4}

17

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Applications of integration: Unit test

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A particle with velocity

v(t)=3 \sqrt{t}

t

is time in seconds, moves in a straight line.

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How far does the particle move from

t=1

t=4

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\square

3

. A particle moves along a lines so that at time

t

0 \leq t \leq \pi

its position is given by

s(t)=-4 \cos t-\frac{t^{2}}{2}+10

. What is the velocity of the particle when its acceleration is zero?

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\cos (t)-1

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\begin{array}{l} s(t)=-4 \cos t-\frac{t^{2}}{2}+10 \\ v(t)=4 \sin (t)-t \\ 0=4 \sin (t)-t \\ t=2.4745768 \\ a(2.47)=4.14 \mathrm{u} / \mathrm{u} \end{array}

6

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With reference to the Figure: What is the rate of acceleration of the object between the

15

20

15

20

seconds)? Note: Please express any deceleration answers by using a negative

(-)

symbol in front of your numerical answer. Note

1

: The units are not required to be included in this instance.

0

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Basic Differentiation Rules and Rates of Change

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121

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05

. Velocity Verify that the average velocity over the time interval

\left[t_{0}-\Delta t, t_{0}+\Delta t\right]

is the same as the instantaneous velocity at

t=t_{0}

for the position function

[In this question

i

j

are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin.] A ship

A

moves with constant velocity

(3i-10j)\,\text{kmh}^{-1}

t

hours, the position vector of

A

r\,\text{km}

t=0

A

is at the point with position vector

(13i+5j)\,\text{km}

j

0

t

In this question

i

j

are horizontal unit vectors. A particle

P

2\,\text{kg}

moves under the action of two forces,

(p i + q j)\,\text{N}

(2q i + p j)\,\text{N}

p

q

are constants. Given that the acceleration of

P

(i - j)\,\text{ms}^{-2}

(a) find the value of

p

and the value of

q

10

. A car has an initial velocity of

300 \mathrm{~m} /

minute. It travels a distance of

0.5 \mathrm{~km}

20

seconds. Use the appropriate formula to calculate the acceleration of the car in

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

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v=u+a t \quad s=u t+\frac{1}{2} a t^{2} \quad v^{2}=u^{2}+2 a s

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a=

constant acceleration,

u=

initial velocity,

v=

final velocity,

s=

displacement fror the position when

t=0

t=

According to Newton's Second Law of Motion, the sum of the forces that act on an object with a mass

m

that moves with an acceleration

a

ma

. An object whose mass is

80

grams has an acceleration of

20

meters per seconds squared. What calculation will give us the sum of the forces that act on the object, in Newtons (which are

\text{kg} \times \text{m/s}^2

Suppose you are driving a car along a straight road. Your velocity at any time

t

(in seconds) is given by the function

v(t)=3 t^{2}-2 t+5

meters per second. Find the total distance traveled by the car from

t=0

t=3

Force equals mass times acceleration.

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\mathrm{F}=\mathrm{ma}

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Solve the equation for the mass,

m

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\mathrm{m}=\frac{[\square]}{[\square]}

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