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Calculus 1 : Acceleration

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Example Questions

Example Question #431 : Spatial Calculus

The displacement of an object at time is defined by the equation $f(t) = \frac{100}{t}$ . What is the acceleration equation for this object?

Possible Answers:

$\frac{-1}{t^3}$

$\frac{100}{t^3}$

$\frac{200}{t^3}$

$\frac{-100}{t^3}$

Correct answer:

$\frac{200}{t^3}$

Explanation:

The acceleration equation is the second derivative of the displacement equation.

$f(t) = \frac{100}{t} = 100t^{-1}$

Therefore the first derivative is equal to

$f'(t) = (-1)(100)t^{-1-1} = -100t^{-2}$

Differentiating a second time gives

$f''(t) = (-2)(-100)t^{-2-1} = 200t^{-3} = \frac{200}{t^3}$

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Example Question #41 : Acceleration

What is the acceleration of an object when its velocity is constantly increasing at a rate of two millimeters per second?

Possible Answers:

$\infty$

$\frac{1}{2}$

Correct answer:

Explanation:

When an object is constantly increasing at a certain rate, it is accelerating at that rate.

Write the equation that represents the object's velocity, where is in seconds.

v(t)=2t

To determine the acceleration, take the derivative of the velocity.

a(t)=2

The acceleration is 2 when the object is constantly increasing at 2 millimeters per second.

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Example Question #432 : Spatial Calculus

The position equation of an object is $f(t) = \sin(t)\cos(t)$ . What is the acceleration of the object at $t=\frac{\pi}{2}$ ?

Possible Answers:

$f''\left(\frac{\pi}{2}\right) = 1$

$f''\left(\frac{\pi}{2}\right) = 2$

$f''\left(\frac{\pi}{2}\right) = -4$

$f''\left(\frac{\pi}{2}\right) = 0$

Correct answer:

$f''\left(\frac{\pi}{2}\right) = 0$

Explanation:

The acceleration of the object is the second derivative of the position equation. To take the derivative of this function, we must use the product rule.

The product rule says

$f(x) = h(x)g(x) \Rightarrow f'(x) = h'(x)g(x) + h(x)g'(x)$ .

From the position equation $f(t) = \sin(t)\cos(t)$ , the first derivative is

$f'(t) = \cos(t)\cos(t) + \sin(t)(-\sin(t)) = \cos^2(t)-\sin^2(t)$ .

To find the second derivative, we must use the chain rule and the power rule.

The chain rule says $f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\times g'(x)$ .

The power rule says that $f(x) = x^n \rightarrow f'(x) = nx^{n-1}$ .

Applying these rules to f'(t) gives,

$f''(t) = 2\cos(t)(-\sin(t)) - 2\sin(t)\cos(t) \Rightarrow \\ \Rightarrow -2\cos(t)\sin(t)-2\sin(t)\cos(t) = -4\sin(t)\cos(t)$

We now use $\frac {\pi}{2}$ as our value for to find our answer.

$f''\left(\frac{\pi}{2}\right) = -4\sin\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi}{2}\right) = -4(1)(0) = 0$

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Example Question #42 : Acceleration

The velocity of an object is given by the equation $v(t) = 7\cos(t)$ . What is the acceleration of the object?

Possible Answers:

$v'(t) = -7\cos(t)\sin(t)$

$v'(t) = -7\cos(t)$

$v'(t) = -7\sin(t)$

$v'(t) = 7\cos(t)\sin(t)$

Correct answer:

$v'(t) = -7\sin(t)$

Explanation:

The acceleration of the object can be found by taking the derivative of the velocity equation.

Using the velocity equation $v(t) = 7\cos(t)$ , the acceleration equation is:

$v'(t) = -7\sin(t)$ .

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Example Question #45 : Acceleration

The velocity of an object is given by the equation v(t) = 4t-3 . What is the acceleration of the object at t = 5 ?

Possible Answers:

Correct answer:

Explanation:

Given the velocity equation v(t) = 4t-3 , the acceleration can be found by differentiating the velocity. We can do this using the power rule, where if

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

This can be done only for terms in the equation that have the variable . All other terms become .

Therefore,

$a(t) = 4(1)t^{(1-1)} - 0 = 4t^0 = 4$ .

Therefore, the acceleration at t = 5 is,

a(5) = 4 .

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Example Question #46 : Acceleration

The position of an object is given by the equation x(t) = 100t^3-1000t^2 . What is the acceleration of the object at t = 10 ?

Possible Answers:

4000

5000

700

12000

Correct answer:

4000

Explanation:

Given the position equation x(t) = 100t^3-1000t^2 , the acceleration of the object can be found by differentiating twice. This can be done by using the power rule where if

$h(x) = x^n \rightarrow h'(x) = nx^{n-1}$ .

Therefore, the first derivative is

$v(t) = 100(3)t^{(3-1)}- 1000(2)t^{(2-1)}= 300t^2 - 2000t$ .

Differentiating again gives us the acceleration equation.

$a(t) = 300(2)t^{(2-1)} - 2000(1)t^{(1-1)}= 600t - 2000$

We can now set time t= 10 to find the acceleration.

a(10) = 600(10) - 2000 = 6000 - 2000 = 4000

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Example Question #434 : Spatial Calculus

The velocity of an object is given by the equation v(t) = 4t^2 - 2 . What is the acceleration of the object at t = 2 ?

Possible Answers:

Correct answer:

Explanation:

The acceleration of the object can be found by differentiating the velocity equation v(t) = 4t^2 -2 .

To differentiate this equation we must use the power rule where

$h(x) = x^n \rightarrow h'(x) = nx^{n-1}$ .

This rule applies to the first term in the velocity equation. The other term does not have the variable and therefore becomes when differentiated.

The derivative of the velocity equation is then,

$a(t) =v'(t) = 4(2)t^{2-1} - 0 = 8t$ .

We now use the value of for to find the acceleration at t = 2 .

a(2) = 8(2) = 16

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Example Question #434 : Spatial Calculus

The position equation of an object is given by the equation $x(t) = t^3 - \sin(t)$ . What is the acceleration equation of the object?

Possible Answers:

$a(t) = 6t - \sin(t)$

$a(t) = 3t - \sin(t)$

$a(t) = 3 - \sin(t)$

$a(t) = 6t + \sin(t)$

Correct answer:

$a(t) = 6t + \sin(t)$

Explanation:

The acceleration equation can be found by differentiating the position equation twice. The first derivative will be the velocity equation, and the second derivative will be the acceleration equation.

To differentiate the position equation we must use the power rule where if

$h(x) = x^n \rightarrow h'(x) = nx^{n-1}$ .

We must also use the fact that when the function $\sin(t)$ is differentiated, it becomes $\cos(t)$ , and when $\cos(t)$ is differentiated it becomes $-\sin(t)$ .

Applying these rules to the position equation gives us,

$x'(t) = v(t) = (3)t^{(3-1)} - (\cos(t)) = 3t^2 - \cos(t)$ .

Differentiating a second time gives,

$a(t) = v'(t) = 3(2)t^{(2-1)} - (-\sin(t)) = 6t + \sin(t)$ .

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Example Question #433 : Calculus

The jerk of an object is given by the equation j(x) = 10t -100 . What is the acceleration of the object at t = 2 , if the initial acceleration is ?

Possible Answers:

Correct answer:

Explanation:

The acceleration of the object can be found by integrating the jerk equation j(t) = a'(t) = 10t -100 . The jerk equation can be rewritten as j(t) = 10t^1 - 100t^0 . To integrate the equation we must use the power rule where

$h'(x) = x^n \rightarrow h(x) = \frac{x^{n+1}}{n+1}$ .

Applying this to the jerk equation gives us,

$a(t) = \int(10t^1 - 100t^0)dt= \frac{10t^{1+1}}{1+1} - \frac{100t^{0+1}}{0+1} + C$

$= \frac{10}{2}t^2 - \frac{100}{1}t^1+C \rightarrow$

$\rightarrow 5t^2 - 100t + C$ .

To complete the equation, we must find the value of using the initial acceleration.

a(0) = 5(0)^2 - 100(0) + C = 5

Therefore C = 5 and the acceleration equation is

a(t) = 5t^2 -100t +5 .

We can now find the acceleration at t = 2 .

a(2) = 5(2)^2 - 100(2) +5 = 20 - 200 +5 = -175

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Example Question #50 : Acceleration

The position of an object is given by the equation $x(t) = \sin(2t) - e^{4t}$ . What is the acceleration of the object?

Possible Answers:

$a(t) = -4\sin(2t) - 16e^{4t}$

$a(t) = -\sin(2t) - e^{4t}$

$a(t) = 4\sin(2t) - 16e^{4t}$

$a(t) = 4\sin(2t) + 16e^{4t}$

Correct answer:

$a(t) = -4\sin(2t) - 16e^{4t}$

Explanation:

The acceleration of the object can be found by differentiating the position equation $x(t) = \sin(2t) - e^{4t}$ twice. To differentiate the position equation, we must use the chain rule where if

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x)) \times g'(x)$ .

To differentiate the position equation we must also know that the derivative of $\sin(t)$ is $\cos(t)$ , the derivative of $\cos(t)$ is $-\sin(t)$ , and the derivative of e^t is e^t . Applying this to each term of the position equation gives us,

$x'(t) = \cos(2t)(2) - e^{4t}(4) = 2\cos(2t) - 4e^{4t}$ .

Differentiating the derivative of the position equation will give us the acceleration equation. To do this we must again use the chain rule.

$a(t) = x''(t) = 2(-\sin(2t))(2) - 4e^{4t}(4) = -4\sin(2t) - 16e^{4t}$

Therefore, the acceleration equation is

$a(t) = -4\sin(2t) - 16e^{4t}$ .

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Calculus 1 : Acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #431 : Spatial Calculus

Example Question #41 : Acceleration

Example Question #432 : Spatial Calculus

Example Question #42 : Acceleration

Example Question #45 : Acceleration

Example Question #46 : Acceleration

Example Question #434 : Spatial Calculus

Example Question #434 : Spatial Calculus

Example Question #433 : Calculus

Example Question #50 : Acceleration

All Calculus 1 Resources

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