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Calculus 1 : How to find acceleration

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Example Questions

Example Question #592 : Calculus

The position of a particle is given by the function $p(t)=2^{cos(t)}$ . Find the acceleration of the particle at time $t=2\pi$

Possible Answers:

3.71

-5.12

-3.77

-1.39

2.01

Correct answer:

-1.39

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 time derivative of the position function.

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

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

To take the derivatives of the function

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

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

Derivative of an exponential:

d[a^u]=a^uduln(a)

Trigonometric derivative:

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

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

Product rule: d[uv]=udv+vdu

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

Using the above properties, the velocity function is

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

And the acceleration function is

$a(t)=-cos(t)2^{cos(t)}ln(2)+sin^2(t)2^{cos(t)}ln^2(2)$

$a(2\pi)=-cos(2\pi)2^{cos(2\pi)}ln(2)+sin^2(2\pi)2^{cos(2\pi)}ln^2(2)$

$a(2\pi)=-1.39$

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Example Question #593 : Calculus

Find the acceleration at time t=1 of a particle whose position function is $p(t)=2t5^{3t}$

Possible Answers:

3778.9

1786.5

11289.1

8242.3

4002.9

Correct answer:

8242.3

Explanation:

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

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

To take the derivatives of the function

$p(t)=2t5^{3t}$

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

Derivative of an exponential:

d[a^u]=a^uduln(a)

Product rule: d[uv]=udv+vdu

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

Using the above properties, the velocity function is

$v(t)=2(5^{3t})+6t5^{3t}ln(5)$

And the acceleration function is

$a(t)=6(5^{3t})ln(5)+6(5^{3t})ln(5)+18t(5^{3t})ln^2(5)$

$a(1)=6(5^{3})ln(5)+6(5^{3})ln(5)+18(5^{3})ln^2(5)$

a(1)=8242.3

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Example Question #594 : Calculus

Find the acceleration of a particle at time t=1 if its position is given by the function $p(t)=2^{ln(t)}$

Possible Answers:

-0.21

0.33

-5.07

3.01

Correct answer:

-0.21

Explanation:

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

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

To take the derivatives of the function

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

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

Derivative of an exponential:

d[a^u]=a^uduln(a)

Derivative of a natural log:

$d[ln(u)]=\frac{du}{u}$

Product rule: d[uv]=udv+vdu

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

Using the above properties, the velocity function is

$v(t)=t^{-1}2^{ln(t)}ln(2)$

And the acceleration function is

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

a(t)=-0.21

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Example Question #201 : How To Find Acceleration

The velocity of a particle is given by the function $v(t)=3^{(t+1)^2}$ . What is the particle's acceleration at time t=1 ?

Possible Answers:

128.9

19.8

6.6

29.7

356.0

Correct answer:

356.0

Explanation:

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

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

To take the derivative of the function

$v(t)=3^{(t+1)^2}$

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

Derivative of an exponential:

d[a^u]=a^uduln(a)

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

Using the above properties, the acceleration function is

$a(t)=2(t+1)(3^{(t+1)^2})ln(3)$

$a(1)=2(1+1)(3^{(1+1)^2})ln(3)$

a(1)=356.0

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Example Question #202 : How To Find Acceleration

The position of a car is given by the function p(t)=t^3-3t^2+6t+5 . At what time does the car cease to accelerate?

Possible Answers:

0.5

1.5

Correct answer:

Explanation:

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

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

Since p(t)=t^3-3t^2+6t+5

Using the above properties, the velocity function is

v(t)=3t^2-6t+6

And the acceleration function is

a(t)=6t-6

The acceleration is zero at time t=1

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Example Question #201 : Acceleration

The position of a particle is given by the function $p(t)=\frac{t^4}{12}-\frac{7t^3}{6}+6t^2+\frac{1}{7}t+\frac{54}{131}$ . At what time does the particle first cease accelerate?

Possible Answers:

Correct answer:

Explanation:

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

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

Since $p(t)=\frac{t^4}{12}-\frac{7t^3}{6}+6t^2+\frac{1}{7}t+\frac{54}{131}$

Using the above properties, the velocity function is

$v(t)=\frac{t^3}{3}-\frac{7t^2}{2}+12t+\frac{1}{7}$

And the acceleration function is

a(t)=t^2-7t+12

To find when the particle is no longer accelerating, set this equation equal to zero:

t^2-7t+12=0

(t-3)(t-4)=0

t=3,4

The particle first has zero acceleration at time t=3

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Example Question #598 : Calculus

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

Possible Answers:

$\frac{\pi}{3}$

$\frac{\pi}{12}$

$\frac{\pi}{2}$

$\frac{\pi}{6}$

$\frac{\pi}{4}$

Correct answer:

$\frac{\pi}{6}$

Explanation:

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

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

To take the derivative of the function

$p(t)=t^2+\cos(\frac{1}{2}t)$

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

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

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

Using the above properties, the velocity function is

$v(t)=2t-2\sin(2t)$

And the acceleration function is

$a(t)=2-4\cos(2t)$

To find when the acceleration is zero, set this equation to zero:

$2-4\cos(2t)=0$

$\cos(2t)=\frac{1}{2}$

$2t=\cos^{-1}(\frac{1}{2})$

$t=\frac{\pi}{6}$

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Example Question #599 : Calculus

The position of a particle is given by the function p(t)=3t^2-e^t at what time does the particle cease to accelerate?

Possible Answers:

1.79

2.24

0.88

6.61

3.14

Correct answer:

1.79

Explanation:

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

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

To take the derivative of the function

p(t)=3t^2-e^t

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

Derivative of an exponential:

d[e^u]=e^udu

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

Using the above properties, the velocity function is

v(t)=6t-e^t

And the acceleration function is

a(t)=6-e^t

Set this equation to zero to find when the particle ceases to accelerate:

6-e^t=0

e^t=6

$t=\ln(6)=1.79$

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Example Question #204 : How To Find Acceleration

The position of a particle is given by the function $x(t)=\frac{t^4}{12}-\frac{5}{6}t^3+3t^2+2t+\frac{1}{8}$ , $y(t)=t^3-9t^2+\frac{5}{14}t+12$ . At what time, if at all, does the particle cease to accelerate?

Possible Answers:

The particle never ceases to accelerate.

Correct answer:

Explanation:

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

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

For the position functions

$x(t)=\frac{t^4}{12}-\frac{5}{6}t^3+3t^2+2t+\frac{1}{8}$

$y(t)=t^3-9t^2+\frac{5}{14}t+12$

Using the above properties, the velocity functions are

$v_x(t)=\frac{t^3}{3}-\frac{5}{2}t^2+6t+2$

$v_y(t)=3t^2-18t+\frac{5}{14}$

And the acceleration functions are

a_x(t)=t^2-5t+6

a_y(t)=6t-18

Now, to find when there is zero acceleration in a direction, set these equations equal to zero:

a_x(t)=t^2-5t+6=(t-3)(t-2)=0

t=2,3

a_y(t)=6t-18=0

t=3

At t=3 , acceleration is zero in both and directions.

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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:

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

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

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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Calculus 1 : How to find acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #592 : Calculus

Example Question #593 : Calculus

Example Question #594 : Calculus

Example Question #201 : How To Find Acceleration

Example Question #202 : How To Find Acceleration

Example Question #201 : Acceleration

Example Question #598 : Calculus

Example Question #599 : Calculus

Example Question #204 : How To Find Acceleration

Example Question #201 : Acceleration

All Calculus 1 Resources

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