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Calculus 1 : Acceleration

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Example Questions

Example Question #552 : Spatial Calculus

Find the acceleration function a(t) if the position function is $s(t) = e^{4t^2} + t$ .

Possible Answers:

$a(t) = 8e^{4t^2}(8t^2 - 1)$

$a(t) = e^{4t^2}(8t^2 + 1)$

$a(t) = 8e^{4t^2}(1 - 8t^2)$

$a(t) = 4e^{4t^2}(8t^2 + 1)$

$a(t) = 8e^{4t^2}(8t^2 + 1)$

Correct answer:

$a(t) = 8e^{4t^2}(8t^2 + 1)$

Explanation:

Acceleration is equal to the second derivative of position with respect to time.

Recall the following rules of differentiation to help solve this problem.

Power Rule:

$\frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}$

Product Rule:

$\frac{\mathrm{d} }{\mathrm{d} t}(u(t)v(t)) = u(t)v'(t)+u'(t)v(t)$

Exponentials Rule:

$\frac{\mathrm{d} }{\mathrm{d} t}(ae^{bt}) = abe^{bt}$

Therefore, by the product rule, the power rule, and the rule for differentiating exponentials,

$\\a(t) = \frac{\mathrm{d} }{\mathrm{d} t}\left(\frac{\mathrm{d} }{\mathrm{d} t}(e^{4t^2} + t)\right)\\ \\ a(t) = \frac{\mathrm{d} }{\mathrm{d} t}(8te^{4t^2}+ 1)\\ \\ a(t) = 64t^2e^{4t^2} + 8e^{4t^2}\\ \\ a(t) = 8e^{4t^2}(8t^2 + 1)$

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Example Question #553 : Spatial Calculus

The position of a snake's head as it moves through the sand is given by $s(t)=3sin\left(\frac{1}{2}t\right)$ . This distance is in meters. At what rate is its head accelerating at $\frac{\pi}{3}$ seconds?

Possible Answers:

1.5m/s^2

-1.5m/s^2

$-\frac{3\sqrt3}{8}m/s^2$

$\frac{3}{8}m/s^2$

$-\frac{3}{8}m/s^2$

Correct answer:

$-\frac{3}{8}m/s^2$

Explanation:

The position is given by the function $s(t)=3sin(\frac{1}{2}t)$ . First find the function for velocity by finding the derivative using the chain rule:

$v(t)=3cos(\frac{1}{2}t)*\frac{1}{2} = 1.5cos(\frac{1}{2}t)$

Then find the function for acceleration by finding the derivative using the chain rule:

$a(t)=1.5*-sin(\frac{1}{2}t)*\frac{1}{2}=-\frac{3}{4}sin(\frac{1}{2}t)$

We are seeking the acceleration at $\frac{\pi}{3}$ seconds, so plug in $\frac{\pi}{3}$ for t:

$a(\frac{\pi}{3})=-\frac{3}{4}sin(\frac{1}{2}*\frac{\pi}{3})= -\frac{3}{4}sin(\frac{\pi}{6})= -\frac{3}{4}*\frac{1}{2}= -\frac{3}{8}$

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Example Question #554 : Spatial Calculus

An object is traveling along a path determined by the function $g(x)=sin(8x)-cos\left(\frac{1}{2}x\right)$ . Use meters for position and seconds for time. What is its initial acceleration?

Possible Answers:

1m/s^2

8m/s^2

0.25m/s^2

-64m/s^2

$-\frac{1}{4}m/s^2$

Correct answer:

0.25m/s^2

Explanation:

Take the first derivative to determine the function for velocity, using the chain rule:

$g'(x)=8cos(8x)+\frac{1}{2}sin(\frac{1}{2}x)$ .

Now use the chain rule again to find the second derivative, the function for acceleration:

$g''(x)=-64sin(8x)+\frac{1}{4}cos(\frac{1}{4}x)$

We're looking for the acceleration at x=0 , so plug in 0:

$g''(0)=-64sin(8*0)+\frac{1}{4}cos(\frac{1}{4}*0)$

At 0, sine is 0 and cosine is 1, so this results in

$g''(0)=0+\frac{1}{4}*1 = 0.25$

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Example Question #555 : Spatial Calculus

Given the velocity function, find the acceleration at t=4 .

v(t)=t^3-3t^2-18t+7

Possible Answers:

None of these

Correct answer:

Explanation:

The derivative of velocity is accleration.

The power rule states that the derivative of x^n is $nx^{n-1}$ .

Thus the acceleration function is

a(t)=3t^2-6t-18 .

The last step is to plug in 4 to the acceleration function.

a(4)=3(4)^2-6(4)-18

=48-24-18

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Example Question #556 : Spatial Calculus

Given the position function, find the acceleration function.

p(t)=t^5+2t^4-5t^3+t^2-10t+15

Possible Answers:

20t^4+24t^3-30t^2+2t-10

None of these

5t^4+8t^3-15t^2+2t-10

20t^3+24t^2-30t+2

20t^3+24t^2-30t

Correct answer:

20t^3+24t^2-30t+2

Explanation:

Velocity is the derivative of position. Acceleration is the derivative of velocity. That means that the second derivative of position is the acceleration.

The derivative of x^n is $nx^{n-1}$ .

So using these rules

v(t)=5t^4+8t^3-15t^2+2t-10

a(t)=20t^3+24t^2-30t+2 .

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Example Question #557 : Spatial Calculus

The position of a particle is given by the function: $x(t)=e^{2t}t^2$

What is its acceleration function?

Possible Answers:

$e^{2t}t^2+e^{2t}t+2e^{2t}$

$4e^{2t}t^2+4e^{2t}t+2e^{2t}$

$4e^{2t}t^2+8e^{2t}t$

$e^{2t}t^2+4e^{2t}t+2e^{2t}$

$4e^{2t}t^2+8e^{2t}t+2e^{2t}$

Correct answer:

$4e^{2t}t^2+8e^{2t}t+2e^{2t}$

Explanation:

Acceleration can be found as the second time derivative of the position function, or the first time derivative of the velocity function:

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

Begin by finding the velocity function by deriving the postion function with respect to time:

$v(t)=2e^{2t}t^2+2e^{2t}t$

After that, derive once more to find acceleration:

$a(t)=4e^{2t}t^2+4e^{2t}t+4e^{2t}t+2e^{2t}$

$a(t)=4e^{2t}t^2+8e^{2t}t+2e^{2t}$

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Example Question #551 : Spatial Calculus

The velocity of a function is given by the function v(t)=sin^2(t^2) . What is the acceleration function?

Possible Answers:

2tsin(2t^2)

2tcos(t^2)

2cos(2t^2)

2sin(2t^2)

2tcos(2t^2)

Correct answer:

2tsin(2t^2)

Explanation:

Acceleration is given as the time derivative of velocity:

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

Therefore, if velocity is given as:

v(t)=sin^2(t^2)

Then acceleration is

a(t)=4tsin(t^2)cos(t^2)

which can also be written as

a(t)=2tsin(2t^2)

If the derivative seems tricky, it may help to view the velocity function as

v(t)=sin(t^2)sin(t^2)

and then to use the product rule of derivatives.

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Example Question #559 : Spatial Calculus

A particle moves with the following position equation:

$f(x)=\cos(e^{x})+x^4+3$

What is the accleration of the particle at the position x=0 ?

Possible Answers:

None of the other answers.

-sin(1)-cos(1)

Correct answer:

-sin(1)-cos(1)

Explanation:

To find the acceleration function, we must find the second derivative of the position equation (which is the acceleration equation):

$f'(x)=-e^x\sin(x)+4x^3$

$f''(x)=-e^x\sin(x)-e^{2x}\cos(e^x)+12x^2$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f' (g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Now, plug in the point x=0 into the second derivative function to find the acceleration of the particle:

f''(0)=a(0)=-sin(1)-cos(1)

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Example Question #560 : Spatial Calculus

A particle's position at any time is given by the function x(t)=t^3 +4t^2-sin(2t) .

What is its acceleration at time t=0 ?

Possible Answers:

Correct answer:

Explanation:

Acceleration is the second time derivative of the positional function and the first time derivative of the velocity function:

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

If position is expressed as:

x(t)=t^3 +4t^2-sin(2t)

Then acceleration can be found to be:

a(t)=6t+8+4sin(2t)

a(0)=8+0=8

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Example Question #561 : Calculus

Find the acceleration function of a vehicle if its velocity function is $v(t)=3t^{2}+15t-7$ .

Possible Answers:

a(t)=6t+15

a(t)=6t

$a(t)=3t^{2}-7$

a(t)=8

$a(t)=3t^{2}+15t$

Correct answer:

a(t)=6t+15

Explanation:

The acceleration function is the first derivative of the velocity function, or

$a(t)=v'(t)=\frac{d}{dt}(v(t))$ .

We will use the power rule, $\frac{d}{dt}(t^{c})=ct^{c-1}$ , where is a constant,

and the constant rule, $\frac{d}{dt}(c)=0$ , where is a constant, to find the derivative of this velocity function.

For this problem:

$a(t)=\frac{d}{dt}(v(t))=3\frac{d}{dt}(t^{2})+15\frac{d}{dt}(t)-\frac{d}{dt}(7)$

a(t)=3*2t+15

a(t)=6t+15

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Calculus 1 : Acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #552 : Spatial Calculus

Example Question #553 : Spatial Calculus

Example Question #554 : Spatial Calculus

Example Question #555 : Spatial Calculus

Example Question #556 : Spatial Calculus

Example Question #557 : Spatial Calculus

Example Question #551 : Spatial Calculus

Example Question #559 : Spatial Calculus

Example Question #560 : Spatial Calculus

Example Question #561 : Calculus

All Calculus 1 Resources

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