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Calculus 1 : Calculus

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #201 : Acceleration

The position of a particle starting at t=0 is given by the function $p(t)=t^3+\sin(t^2)$ . At what time does the particle first cease to accelerate?

Possible Answers:

-3.49

1.29

-2.39

-2.02

2.64

Correct answer:

2.64

Explanation:

Acceleration of a particle can be found by taking the derivative of the velocity function with respect to time. Recall that a derivative gives the rate of change of some parameter, relative to the change of some other variable. When we take the derivative of velocity with respect to time, we are evaluating how velocity changes over time; i.e acceleration! This is just like finding velocity by taking the derivative of the position function.

$a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$

To take the derivative of the function

$p(t)=t^3+\sin(t^2)$

We'll need to make use of the following derivative rule(s):

$d[\sin(u)]=\cos(u)du$

$d[\cos(u)]=-\sin(u)du$

Product rule: d[uv]=udv+vdu

Note that and may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$v(t)=3t^2+2t\cos(t^2)$

And the acceleration function is

$a(t)=6t+2\cos(t^2)-4t^2\sin(t^2)$

Set this equation equal to zero to find when the particle ceases to decelerate:

$6t+2\cos(t^2)-4t^2\sin(t^2)=0$

t=-3.49,-3.16,-2.39,-2.02,-0.32,2.64,2.99,3.61

Since we do not consider times earlier than t=0 , the particle first has zero acceleration at time t=2.64

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Example Question #211 : How To Find Acceleration

The position of a particle is given by the function $p(t)=3^{t^2}$ . What is the acceleration of the particle at time t=1.2 ?

Possible Answers:

22.12

5.67

87.91

13.89

44.51

Correct answer:

44.51

Explanation:

$a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$

To take the derivative of the function

$p(t)=3^{t^2}$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

$d[a^u]=a^udu\ln(a)$

Product rule: d[uv]=udv+vdu

Note that and may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$v(t)=2t(3^{t^2})\ln(3)$

And the acceleration function is

$a(t)=2(3^{t^2})\ln(3)+4t^2(3^{t^2})(\ln(3))^2$

At t=1.2

$a(1.2)=2(3^{1.2^2})\ln(3)+4(1.2)^2(3^{1.2^2})(\ln(3))^2=44.51$

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Example Question #212 : Acceleration

The position of a particle is given by the function $p(t)=\cos(2^{3t})$ . What is the particle's acceleration at time t=1.4 ?

Possible Answers:

2934.5

21.2

-1266.0

-830.5

117.88

Correct answer:

-1266.0

Explanation:

$a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$

To take the derivative of the function

$p(t)=\cos(2^{3t})$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

$d[a^u]=a^udu\ln(a)$

Trigonometric derivative:

$d[sin(u)]=\cos(u)du$

$d[cos(u)]=-\sin(u)du$

Product rule: d[uv]=udv+vdu

Note that and may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$v(t)=-3\ln(2)(2^{3t})\sin(2^{3t})$

And the acceleration function is

$a(t)=-9(\ln(2))^2(2^{3t})\sin(2^{3t})-9(\ln(2))^2(2^{3t})^2\cos(2^{3t})$

At t=1.4

$a(1.4)=-9(\ln(2))^2(2^{3(1.4)})\sin(2^{3(1.4)})-9(\ln(2))^2(2^{3(1.4)})^2\cos(2^{3(1.4)})$

a(1.4)=-1266.0

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Example Question #213 : Acceleration

The velocity of a particle is given by the function $v(t)=1.8^{\cos(2.2t)}$ . What is the acceleration of the particle at time t=3.1 ?

Possible Answers:

-1.10

-4.92

-7.17

-2.33

0.83

Correct answer:

-1.10

Explanation:

$a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$

To take the derivative of the function

$v(t)=1.8^{\cos(2.2t)}$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

$d[a^u]=a^udu\ln(a)$

Trigonometric derivative:

$d[\cos(u)]=-\sin(u)du$

Note that may represent large functions, and not just individual variables!

Using the above properties, the acceleration function is

$a(t)=-2.2\sin(2.2t)(1.8^{\cos(2.2t)})\ln(1.8)$

At t=3.1

$a(3.1)=-2.2\sin(2.2(3.1))(1.8^{\cos(2.2(3.1))})\ln(1.8)$

a(3.1)=-1.10

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Example Question #601 : Spatial Calculus

The velocity of a particle is given by the function $v(t)=1.4^{t^2+t^3}$ . What is the particle's accleration at time t=2 ?

Possible Answers:

907.1

228.9

19.1 56.7

305.2

Correct answer:

305.2

Explanation:

$a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$

To take the derivative of the function

$v(t)=1.4^{t^2+t^3}$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

$d[a^u]=a^udu\ln(a)$

Note that may represent large functions, and not just individual variables!

Using the above properties, the acceleration function is

$a(t)=(2t+3t^2)(1.4^{t^2+t^3})\ln(1.4)$

At time t=2

$a(2)=(2(2)+3(2)^2)(1.4^{2^2+2^3})\ln(1.4)=305.2$

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Example Question #215 : Acceleration

The position of a particle is given by the function $p(t)=2^{t^2}-3^{t}$ . What is the particle's acceleration at time t=1.5 ?

Possible Answers:

63.3

20.9

41.2

5.8

13.7

Correct answer:

20.9

Explanation:

$a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$

To take the derivative of the function

$p(t)=2^{t^2}-3^{t}$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

$d[a^u]=a^udu\ln(a)$

Product rule: d[uv]=udv+vdu

Note that and may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

$v(t)=2t(2^{t^2})\ln(2)-3^{t}\ln(3)$

And the acceleration function is

$a(t)=2(2^{t^2})\ln(2)+4t^2(2^{t^2})(\ln(2))^2-3^{t}(\ln(3))^2$

At time t=1.5

$a(1.5)=2(2^{1.5^2})\ln(2)+4(1.5)^2(2^{1.5^2})(\ln(2))^2-3^{1.5}(\ln(3))^2$

a(1.5)=20.9

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Example Question #216 : Acceleration

The velocity of a particle is given by the function $v(t)=\frac{8}{3^{t^2}}$ What is the particle's velocity at time t=2 ?

Possible Answers:

-0.17

0.13

0.22

0.58

-0.43

Correct answer:

-0.43

Explanation:

$a(t)=\frac{d}{dt}v(t)$

$v(t)=\frac{d}{dt}p(t)$

To take the derivative of the function

$v(t)=\frac{8}{3^{t^2}}$ it may help to rewrite it as $v(t)=8(3^{-t^2})$

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

$d[a^u]=a^udu\ln(a)$

Note that may represent large functions, and not just individual variables!

Using the above properties, the acceleration function is

$a(t)=-2t(8)(3^{-t^2})\ln(3)=-16t(3^{-t^2})\ln(3)$

At time t=2

$a(2)=-16(2)(3^{-(2)^2})\ln(3)=-0.43$

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Example Question #217 : Acceleration

A toy car is thrown straight upward into the air. The equation of the position of the object is:

$y(t) = t^{2} - 5t$

What is the instantaneous acceleration of the toy car at any time?

Possible Answers:

$1 \tfrac{m}{s^{2}}$

$4 \tfrac{m}{s^{2}}$

$5 \tfrac{m}{s^{2}}$

$3 \tfrac{m}{s^{2}}$

$2 \tfrac{m}{s^{2}}$

Correct answer:

$2 \tfrac{m}{s^{2}}$

Explanation:

To find the velocity of the toy car, we take the derivative of the position equation. The velocity equation is

v(t) = 2t - 5

Then to find the acceleration of the toy car, we take the derivative again. The acceleration equation is

a(t) = 2

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Example Question #602 : Spatial Calculus

A Spaceship is traveling through the galaxy. The distance traveled by the spaceship over a certain amount of time can be calculated by the equation

$x(t) = 2\frac{m}{s^{2}} (t^{2}) - 10\tfrac{m}{s}(t) + 10 m$

where is the distance traveled in meters and is time in seconds .

What is the instantaneous acceleration of the spaceship at T = 5s

Possible Answers:

$4\tfrac{m}{s^{2}}$

$40\tfrac{m}{s^{2}}$

$200\tfrac{m}{s^{2}}$

$0\tfrac{m}{s^{2}}$

$10\tfrac{m}{s^{2}}$

Correct answer:

$4\tfrac{m}{s^{2}}$

Explanation:

We can find the acceleration of the spaceship over a time frame by taking the derivative of the velocity equation or by taking the derivative of the position equation twice.

The derivative of the position equation is:

$v(t) = 4\tfrac{m}{s^{2}}(t) - 10\tfrac{m}{s}$

The derivative of the velocity equation is:

$a(t) = 4\tfrac{m}{s^{2}}$

The question asked what is the instantaneous acceleration of the spaceship at the $5\:sec$ mark. When we insert $5\:s$ into the acceleration equation, we get $4\tfrac{m}{s^{2}}$ .

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Example Question #211 : Acceleration

A ball was launched across the Mississippi river. The position of the ball as it is traveling across the river is

$x(t) = 2 t^{2} + 10t + 20$ (where is in meters and is in seconds )

What is the instantaneous acceleration of the rock at t = 10 seconds ?

Possible Answers:

$30 \tfrac{m}{s^{2}}$

$4 \tfrac{m}{s^{2}}$

$20 \tfrac{m}{s^{2}}$

$10 \tfrac{m}{s^{2}}$

$1\tfrac{m}{s^{2}}$

Correct answer:

$4 \tfrac{m}{s^{2}}$

Explanation:

To find the velocity of the rock, we take the derivative of the position equation. The velocity equation is

v(t) = 4t + 10

From the velocity equation, we take the derivative again to find the acceleration. The acceleration equation is

a(t) = 4

From the acceleraton equation, we insert t = 10 seconds to find the instantaneous acceleration.

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #201 : Acceleration

Example Question #211 : How To Find Acceleration

Example Question #212 : Acceleration

Example Question #213 : Acceleration

Example Question #601 : Spatial Calculus

Example Question #215 : Acceleration

Example Question #216 : Acceleration

Example Question #217 : Acceleration

Example Question #602 : Spatial Calculus

Example Question #211 : Acceleration

All Calculus 1 Resources

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