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Calculus 1 : How to find acceleration

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Example Questions

Example Question #522 : Spatial Calculus

Given that the velocity of a particle is $\small v(t)=2t\cos t$ , what is the acceleration of the particle at time $\small t=\frac{\pi}{2}$ ?

Possible Answers:

$\small \pi$

$\small -\pi$

$\small -2$

$\small 0$

Correct answer:

$\small -\pi$

Explanation:

Given the velocity $\small v(t)=2t\cos t$ , we know that the acceleration function is given by the derivative of the velocity:

$\small a(t)=v'(t)=2\cos t-2t\sin t$

So now we plug in $t=\frac{\pi}{2}$ to get

$\small a(\pi/2)=2\cos \pi/2-2(\pi/2)\sin \pi/2=-\pi$

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Example Question #523 : Spatial Calculus

A car's position can be represented by the equation x(t)=t^4+3t^2-7 . Given this information, find the car's acceleration when t=2 .

Possible Answers:

Correct answer:

Explanation:

The given equation is used to find position. Based on what we know about derivatives, if we take the derivative of position with respect to time we will be finding the change in position over time, which is velocity. So the first step is to take the derivative. We will get

x'(t)=v(t)=4t^3+6t

to represent our velocity. Next, if we take the derivative of the velocity, we get change in velocity, which is an objects acceleration.

v'(t)=a(t)=12t^2+6

you plug in the given t value of t=2 to find the acceleration at this step in time.

a(2)=12(2)^2+6=54

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Example Question #524 : Spatial Calculus

A car is traveling with a velocity represnted by the function v(t)=3t^2+4 when $t\geq0$ . Using this information what is the vehicle's acceleration when t=3 .

Possible Answers:

Correct answer:

Explanation:

If we take the derivative of the velocity, we get change in velocity with respect to time, which is an objects acceleration.

$\frac{\mathrm{d} }{\mathrm{d} t}v(t)=a(t)=6t$

you plug in the given t value of t=3 to find the acceleration at this step in time.

a(3)=6(3)=18 .

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Example Question #525 : Spatial Calculus

An object's position is represented by the function x(t)=t^4 when $t\geq0$ . Using this information, find the object's acceleration when t=5 .

Possible Answers:

300

500

100

Correct answer:

300

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} t}x(t)=v(t)=4t^3$

to represent our velocity. Next, if we take the derivative of the velocity, we get change in velocity, which is an objects acceleration.

$\frac{\mathrm{d} }{\mathrm{d} t}v(t)=a(t)=12t^2$

you plug in the given t value of t=5 to find the acceleration at this step in time.

a(5)=12(5)^2=300 .

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Example Question #526 : Spatial Calculus

An object's motion is represented by the polynomial x(t)=2t^3+t^2-4t+12 when $t\geq0$ . Using this information, find the objects acceleration when t=1 .

Possible Answers:

Correct answer:

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} t}x(t)=v(t)=6t^2+2t-4$

to represent our velocity. Next, if we take the derivative of the velocity, we get change in velocity, which is an objects acceleration.

$\frac{\mathrm{d} }{\mathrm{d} t}v(t)=a(t)=12t+2$

you plug in the given t value of t=1 to find the acceleration at this step in time.

a(1)=12(1)+2=14 .

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Example Question #522 : Calculus

What is the acceleration function of a particle whose position function is given with respect to time as x(t)=cos(2t)ln(3t) ?

Possible Answers:

$4cos(2t)ln(3t)+4\frac{sin(2t)}{t}-\frac{cos(2t)}{t^2}$

$-4cos(2t)ln(3t)+4\frac{sin(2t)}{t}+\frac{cos(2t)}{t^2}$

$-4cos(2t)ln(3t)-4\frac{sin(2t)}{t}-\frac{cos(2t)}{t^2}$

$4cos(2t)ln(3t)-4\frac{sin(2t)}{t}-\frac{cos(2t)}{t^2}$

$4cos(2t)ln(3t)-4\frac{sin(2t)}{t}+\frac{cos(2t)}{t^2}$

Correct answer:

$-4cos(2t)ln(3t)-4\frac{sin(2t)}{t}-\frac{cos(2t)}{t^2}$

Explanation:

Acceleration is given as the second derivative with respect to time of position:

$a(t)=\frac{d^2}{dt^2}x(t)=\frac{d^2}{dt^2}(cos(2t)ln(3t))$

To take the first derivative, it will be necessary to use the product rule of derivation:

$\frac{d}{dt}uv=\frac{du}{dt}v+\frac{dv}{dt}u$

This yields the velocity function:

$v(t)=-2sin(2t)ln(3t)+3\frac{cos(2t)}{3t}=-2sin(2t)ln(3t)+\frac{cos(2t)}{t}$

Now, to find the acceleration function, we derive once more:

$a(t)=-4cos(2t)ln(3t)-6\frac{sin(2t)}{3t}-2\frac{sin(2t)}{t}-\frac{cos(2t)}{t^2}$

$a(t)=-4cos(2t)ln(3t)-4\frac{sin(2t)}{t}-\frac{cos(2t)}{t^2}$

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Example Question #527 : Spatial Calculus

The velocity of a function with respect to time is given as $v(t)=\frac{t^2}{ln(t)}$ . Determine its acceleration function.

Possible Answers:

$\frac{t(1+2ln(t))}{ln(t)}$

$\frac{2t}{ln(t)}$

2t(1+tln(t))

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

Correct answer:

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

Explanation:

Acceleration is given as the derivative of the velocity function with respect to time:

$a(t)=\frac{d}{dt}v(t)=\frac{d}{dt}(\frac{t^2}{ln(t)})$

To find this derivative, it would be useful to utilize the rule of derivatves for quotients:

$\frac{d}{dt}\frac{u}{v}=\frac{vdu-udv}{v^2}$

$a(t)=\frac{2tln(t)-t^2(\frac{1}{t})}{(ln(t))^2}=\frac{t(2ln(t)-1)}{ln(t)^2}$

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Example Question #528 : Spatial Calculus

If f(x) models the position of a frisbee as a function of time. Find the equation which models the acceleration of the frisbee.

f(x)=x^3+5x^2-4x

Possible Answers:

a(x)=6x+10

a(x)=3x^2+10x-4

a(x)=6

a(x)=-4

Correct answer:

a(x)=6x+10

Explanation:

Recall that velocity if the first derivative of position, and acceleration is the first derivative of velocity. To find the equation which models the acceleration, we need to take the second derivative of f(x).

Recall that for any polynomial we can take the derivative as follows:

f(x)=x^n

$f'(x)=nx^{n-1}$

So our original function:

f(x)=x^3+5x^2-4x

Becomes:

f'(x)=3x^2+10x-4

And then:

f''(x)=6x+10

So our answer is

a(x)=6x+10

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Example Question #529 : Spatial Calculus

If f(x) models the position of a frisbee as a function of time, find the acceleration of the frisbee after seconds.

f(x)=x^3+5x^2-4x

Possible Answers:

$28\frac{m}{s^2}$

$-28\frac{m}{s^2}$

$2.8\frac{m}{s^2}$

$48\frac{m}{s^2}$

Correct answer:

$28\frac{m}{s^2}$

Explanation:

Recall that velocity if the first derivative of position, and acceleration is the first derivative of velocity. To find the acceleration after 3 seconds, we need to find f''(3).

Recall that for any polynomial we can take the derivative as follows:

f(x)=x^n

$f'(x)=nx^{n-1}$

So our original function:

f(x)=x^3+5x^2-4x

Becomes:

f'(x)=3x^2+10x-4

And then:

f''(x)=6x+10

Finally, we need f''(3)

$f''(3)=(6\cdot 3)+10=28\frac{m}{s^2}$

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Example Question #131 : How To Find Acceleration

V(t) models the velocity of a spaceship. Find a function to model the spaceship's acceleration as a function of time.

V(t)=8t^3-3t^2+5

Possible Answers:

A(t)=24t^2-6t

A(t)=24t^2-6t +5

A(t)=48t-6

A(t)=2t^4-t^3+5t

Correct answer:

A(t)=24t^2-6t

Explanation:

V(t) models the velocity of a spaceship. Find a function to model the spaceship's acceleration as a function of time.

Remember how position/velocity/acceleration are all related: Velocity is the first derivative of position, and acceleration is the derivative of velocity.

So, we need to take the derivative of V(t). To find that, we will decrease each exponent by 1 and multiply by the original number:

$V'(t)=3\cdot 8t^{3-1}-2\cdot 3t^{2-1}$

Thus, we get:

V'(t)=A(t)=24t^2-6t

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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 #522 : Spatial Calculus

Example Question #523 : Spatial Calculus

Example Question #524 : Spatial Calculus

Example Question #525 : Spatial Calculus

Example Question #526 : Spatial Calculus

Example Question #522 : Calculus

Example Question #527 : Spatial Calculus

Example Question #528 : Spatial Calculus

Example Question #529 : Spatial Calculus

Example Question #131 : How To Find Acceleration

All Calculus 1 Resources

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