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Calculus 1 : How to find acceleration

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Example Question #61 : How To Find Acceleration

The velocity of an object is given by the equation $v(t) = e^{4t^2}$ . What is the acceleration of the object at time t = 2 ?

Possible Answers:

$4e^{16}$

$8e^{16}$

$16e^{16}$

$e^{16}$

Correct answer:

$16e^{16}$

Explanation:

The acceleration of the object can be found by differentiating the velocity equation. To do this we can use the power rule and the chain rule where if

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

and where if

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

Using these two rules we find the acceleration equation to be

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

We can now solve for the acceleration at time t= 2

$a(2) = 8(2)e^{(4(2)^2)} = 16e^{16}$

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Example Question #62 : How To Find Acceleration

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

Possible Answers:

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

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

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

$a(t) = \frac{t\sin(t)+\cos(t)}{t^2}$

Correct answer:

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

Explanation:

The acceleration of the object can be found by differentiating the velocity equation. To accurately differentiate the velocity equation, the quotient rule can be used where if

$f(x) = \frac{m(x)}{z(x)} \rightarrow f'(x) = \frac{m'(x)z(x) - m(x)z'(x)}{[z(x)]^2}$

Using this equation with the velocity equation gives us

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

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Example Question #63 : How To Find Acceleration

The velocity of an object is given by the equation $v(t) = \frac{-\sin(t)}{7t}$ . What is the acceleration of the object at time t= 1 ?

Possible Answers:

$a(t) = \frac{-7\cos(1)+7\sin(1)}{49}$

$a(t) = \frac{7\cos(1)+7\sin(1)}{49}$

$a(t) = \frac{7\cos(1)-7\sin(1)}{49}$

$a(t) = \frac{7\cos(1)-7\sin(1)}{7}$

Correct answer:

$a(t) = \frac{-7\cos(1)+7\sin(1)}{49}$

Explanation:

The acceleration of the object can be found by differentiating the velocity equation. The velocity equation can be accurately differentiated using the quotient rule and the power rule where if

$f(x) = \frac{m(x)}{z(x)} \rightarrow f'(x) = \frac{m'(x)z(x) - m(x)z'(x)}{[z(x)]^2}$

and if

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

Using these two rules we find that the acceleration equation is

$a(t) = \frac{-\cos(t)(7t)+7(\sin(t))}{49t^2} = \frac{-7t\cos(t)+7\sin(t)}{49t^2}$

We can now find the acceleration at t = 1

$a(1) = \frac{-7(1)\cos(1)+7\sin(1)}{49(1)^2} = \frac{-7\cos(1)+7\sin(1)}{49}$

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Example Question #64 : How To Find Acceleration

The velocity of an object is given by the equation $v(t) = 7e^{3t}\cos(t)$ . What is the acceleration of the object?

Possible Answers:

$a(t) = 7te^{3t}\cos(t)+7e^{3t}\sin(t)$

$a(t) = 7e^{3t}\cos(t)-7e^{3t}\sin(t)$

$a(t) = 21e^{3t}\cos(t)-7e^{3t}\sin(t)$

$a(t) = -21e^{3t}\cos(t)\sin(t)$

Correct answer:

$a(t) = 21e^{3t}\cos(t)-7e^{3t}\sin(t)$

Explanation:

The acceleration of the object is the derivative of the velocity of the object. To differentiate the velocity equation we can use the power rule, chain rule, and product rule. The power rule is where if

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

The chain rule is where if

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

Lastly, The product rule is where if

$f(x)= j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)$

Using these rules the acceleration equation is

$\\a(t) = v'(t) = 7e^{3t}(3(1)t^{(1-1)})(\cos(t)) + 7e^{3t}(-\sin(t)) \\ \\ a(t) = 21e^{3t}\cos(t)-7e^{3t}\sin(t)$

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Example Question #65 : How To Find Acceleration

The position, , of a particle is given by the following equation.

$s(t) = 0.5t^3+sin\left(\frac{t}{2}\right)$

(Use radians for any trig functions)

What is the acceleration of the partice when $t=\pi$ ?

Possible Answers:

$3\pi$

$3\pi+\frac{\sqrt{2}}{2}$

$3\pi-\sqrt{2}$

$3\pi-\frac{1}{4}$

$3\pi-\frac{\sqrt{2}}{4}$

Correct answer:

$3\pi-\frac{1}{4}$

Explanation:

The acceleration, a, of a particle is given by

$a=\frac{d^2s}{dt^2}$

Thus using the power rule and the chain rule we can find our acceleration.

Taking the derivative once we get the velocity.

$\frac{ds}{dt}=1.5t^2+\frac{cos(\frac{t}{2})}{2}$

Taking the derivative a second time we get the acceleration.

$\frac{d^2s}{dt^2}=3t-\frac{sin(\frac{t}{2})}{4}$

Plugging in at $t=\pi$ we find

$\frac{d^2s}{dt^2}=3\pi-\frac{1}{4}$ .

Where we used the fact that $sin\left(\frac{\pi}{2}\right)=1$ .

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Example Question #66 : How To Find Acceleration

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

Possible Answers:

$a(t) = \sin^2(t)\cos^2(t)$

$a(t) = \sin^2(t) + \cos^2(t)$

$a(t) = \sin(t)\cos(t)$

$a(t) = \cos^2(t) - \sin^2(t)$

Correct answer:

$a(t) = \cos^2(t) - \sin^2(t)$

Explanation:

The acceleration of the object is equal to the derivative of the object's velocity. To accurately differentiate the the velocity of the object we must use the product rule where if

$f(x)= j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)$ .

Using this rule we find the acceleration to be

$a(t) = (\cos(t))\cos(t) + \sin(t)(-\sin(t)) = \cos^2(t) - \sin^2(t)$ .

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Example Question #67 : How To Find Acceleration

The position of an object is given by the equation x(t) = 4t^3 - 7t^2 . What is the acceleration of the object?

Possible Answers:

a(t) = 24t +14

a(t) = 4

a(t) = 24

$a(t) = \frac{t^5}{5} +\frac{7t^4}{12}$

Correct answer:

a(t) = 24t +14

Explanation:

The acceleration of the object can be found by differentiating the position of the object twice. This can be done using the power rule where if

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

Therefore the velocity of the object is

$v(t) = 4(3)t^{3-1} + 7(2)t^{2-1} = 12t^2 + 14t$ .

This is repeated to find the acceleration of the object

$a(t) = 12(2)t^{2-1} + 14(1)t^{1-1} = 24t + 14$ .

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Example Question #68 : How To Find Acceleration

The velocity of an object is given by the equation $v(t) = e^{\cos(2t)}$ . What is the acceleration of the object at time $t= \frac{\pi}{4}$ ?

Possible Answers:

None of the above.

Correct answer:

Explanation:

The acceleration of the object can be found by differentiating the velocity. To differentiate the velocity accurately the chain rule and power rule can be used where if

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

and if

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

Differentiating the velocity equation we get,

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

We can now solve for the acceleration at $t= \frac{\pi}{4}$

$a\left(\frac{\pi}{4}\right) = -2\sin\left(2\left(\frac{\pi}{4}\right)\right)e^{\cos(2(\frac{\pi}{4}))} = -2\sin\left(\frac{\pi}{2}\right)e^{\cos(\frac{\pi}{2})} = -2(1)e^{(0)} = -2$

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Example Question #69 : How To Find Acceleration

The velocity of an object is $v(t) = \cos^2(t)\sin^2(t)$ . What is the acceleration of the object?

Possible Answers:

$a(t) = -2\cos(t)\sin^3(t) - 2\sin(t)\cos^3(t)$

$a(t) = 2\cos(t)\sin^3(t) + 2\sin(t)\cos^3(t)$

a(t) = 0

$a(t) = -2\cos(t)\sin^3(t) + 2\sin(t)\cos^3(t)$

Correct answer:

$a(t) = -2\cos(t)\sin^3(t) + 2\sin(t)\cos^3(t)$

Explanation:

The acceleration can be found by differentiating the velocity of the object. This can be done accurately using the power rule, chain rule, and product rule where if

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

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

and if

$f(x)= j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)$ .

Therefore the acceleration of the object is

$\\a(t) = (2)\cos^{(2-1)}(t)(-\sin(t))(\sin^2(t))+\cos^2(t)((2)\sin^{(2-1)}(t))(\cos(t)) \\ \Rightarrow -2\cos(t)\sin^3(t) + 2\sin(t)\cos^3(t)$ .

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Example Question #70 : How To Find Acceleration

The jerk on an object is j(t) = 11t . What is the acceleration on the object if its initial acceleration is ?

Possible Answers:

a(t) = 11t^2 - 10

a(t) = 11

$a(t) = \frac{11t^2}{2} - 10$

$a(t) = \frac{11t^2}{2} + 10$

Correct answer:

$a(t) = \frac{11t^2}{2} + 10$

Explanation:

The object acceleration can be found by integrating the object's jerk. This can be done using the power rule where

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

Therefore the acceleration of the object is

$a(t) = \int (11t)dt = \frac{11t^{(1+1)}}{1+1} +C = \frac{11t^2}{2} +C$

Solving for the constant C using the initial condition given in the question is as follows.

$a(0) = \frac{11(0)^2}{2} +C = 0 +C = 10$

Therefore C =10 and $a(t) = \frac{11t^2}{2} + 10$ .

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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 #61 : How To Find Acceleration

Example Question #62 : How To Find Acceleration

Example Question #63 : How To Find Acceleration

Example Question #64 : How To Find Acceleration

Example Question #65 : How To Find Acceleration

Example Question #66 : How To Find Acceleration

Example Question #67 : How To Find Acceleration

Example Question #68 : How To Find Acceleration

Example Question #69 : How To Find Acceleration

Example Question #70 : How To Find Acceleration

All Calculus 1 Resources

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