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Calculus 1 : How to find acceleration

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Example Questions

Example Question #211 : How To Find Acceleration

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

Possible Answers:

22.12 $\displaystyle 22.12$

5.67 $\displaystyle 5.67$

87.91 $\displaystyle 87.91$

13.89 $\displaystyle 13.89$

44.51 $\displaystyle 44.51$

Correct answer:

44.51 $\displaystyle 44.51$

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.

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

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

To take the derivative of the function

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

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

Derivative of an exponential:

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

Product rule: $\displaystyle d[uv]=udv+vdu$

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

Using the above properties, the velocity function is

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

And the acceleration function is

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

At t=1.2 $\displaystyle t=1.2$

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

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Example Question #212 : Acceleration

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

Possible Answers:

2934.5 $\displaystyle 2934.5$

21.2 $\displaystyle 21.2$

-1266.0 $\displaystyle -1266.0$

-830.5 $\displaystyle -830.5$

117.88 $\displaystyle 117.88$

Correct answer:

-1266.0 $\displaystyle -1266.0$

Explanation:

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

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

To take the derivative of the function

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

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

Derivative of an exponential:

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

Trigonometric derivative:

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

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

Product rule: $\displaystyle d[uv]=udv+vdu$

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

Using the above properties, the velocity function is

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

And the acceleration function is

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

At t=1.4 $\displaystyle t=1.4$

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

$\displaystyle a(1.4)=-1266.0$

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Example Question #213 : Acceleration

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

Possible Answers:

-1.10 $\displaystyle -1.10$

-4.92 $\displaystyle -4.92$

-7.17 $\displaystyle -7.17$

-2.33 $\displaystyle -2.33$

0.83 $\displaystyle 0.83$

Correct answer:

-1.10 $\displaystyle -1.10$

Explanation:

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

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

To take the derivative of the function

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

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

Derivative of an exponential:

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

Trigonometric derivative:

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

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

Using the above properties, the acceleration function is

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

At t=3.1 $\displaystyle t=3.1$

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

$\displaystyle a(3.1)=-1.10$

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Example Question #601 : Spatial Calculus

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

Possible Answers:

228.9 $\displaystyle 228.9$

305.2 $\displaystyle 305.2$

907.1 $\displaystyle 907.1$

19.1 $\displaystyle 19.1$ 56.7 $\displaystyle 56.7$

Correct answer:

305.2 $\displaystyle 305.2$

Explanation:

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

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

To take the derivative of the function

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

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

Derivative of an exponential:

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

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

Using the above properties, the acceleration function is

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

At time t=2 $\displaystyle t=2$

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

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Example Question #215 : Acceleration

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

Possible Answers:

63.3 $\displaystyle 63.3$

20.9 $\displaystyle 20.9$

41.2 $\displaystyle 41.2$

5.8 $\displaystyle 5.8$

13.7 $\displaystyle 13.7$

Correct answer:

20.9 $\displaystyle 20.9$

Explanation:

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

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

To take the derivative of the function

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

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

Derivative of an exponential:

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

Product rule: $\displaystyle d[uv]=udv+vdu$

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

Using the above properties, the velocity function is

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

And the acceleration function is

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

At time t=1.5 $\displaystyle t=1.5$

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

$\displaystyle a(1.5)=20.9$

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Example Question #216 : Acceleration

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

Possible Answers:

-0.17 $\displaystyle -0.17$

0.13 $\displaystyle 0.13$

0.22 $\displaystyle 0.22$

0.58 $\displaystyle 0.58$

-0.43 $\displaystyle -0.43$

Correct answer:

-0.43 $\displaystyle -0.43$

Explanation:

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

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

To take the derivative of the function

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

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

Derivative of an exponential:

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

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

Using the above properties, the acceleration function is

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

At time t=2 $\displaystyle t=2$

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

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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:

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

$\displaystyle v(t) = 2t - 5$

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

$\displaystyle a(t) = 2$

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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

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

where $\displaystyle x$ is the distance traveled in meters and $\displaystyle t$ is time in seconds $\displaystyle seconds$.

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

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

The derivative of the velocity equation is:

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

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

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Example Question #602 : Spatial Calculus

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

$\displaystyle v(t) = 4t + 10$

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

$\displaystyle a(t) = 4$

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

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Example Question #611 : Spatial Calculus

A driver is speeding through the interstate. The distance traveled by the driver over a certain amount of time can be calculated by the equation

$x(t) = 20\frac{m}{s^{2}} (t^{2}) + 10m$ $\displaystyle x(t) = 20\frac{m}{s^{2}} (t^{2}) + 10m$

where $\displaystyle x$ is the distance traveled in meters $\displaystyle meters$ and $\displaystyle t$ is time in seconds $\displaystyle seconds$.

What is the equation for the acceleration of the driver?

Possible Answers:

$a(t) = 20\tfrac{m}{s} (t)$ $\displaystyle a(t) = 20\tfrac{m}{s} (t)$

$a(t) = 20\tfrac{m}{s} (t^{2})$ $\displaystyle a(t) = 20\tfrac{m}{s} (t^{2})$

$a(t) = 20\tfrac{m}{s^{2}}$ $\displaystyle a(t) = 20\tfrac{m}{s^{2}}$

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

$a(t) = 40\tfrac{m}{s} (t)$ $\displaystyle a(t) = 40\tfrac{m}{s} (t)$

Correct answer:

$a(t) = 20\tfrac{m}{s^{2}}$ $\displaystyle a(t) = 20\tfrac{m}{s^{2}}$

Explanation:

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

$v(t) = 20\tfrac{m}{s} (t)$ $\displaystyle v(t) = 20\tfrac{m}{s} (t)$

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

$a(t) = 20\tfrac{m}{s}$ $\displaystyle a(t) = 20\tfrac{m}{s}$

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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 #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 #602 : Spatial Calculus

Example Question #611 : Spatial Calculus

All Calculus 1 Resources

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