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Calculus 1 : How to find acceleration

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Example Questions

Example Question #471 : Calculus

The position of an object is given by the equation $x(t) = e^{(2t)}\sin(3t)$ . What is the acceleration of the object?

Possible Answers:

$a(t) =12e^{(2t)}\cos(3t)-5e^{(2t)}\sin(3t)$

$a(t) = 4e^{(2t)}\sin(3t)+6e^{(2t)}\cos(3t)-9e^{(2t)}\sin(3t)$

$a(t) = 4e^{(2t)}\sin(3t)+6e^{(2t)}\cos(3t)+3e^{(2t)}\sin(3t)$

$a(t) = 10e^{(2t)}\sin(3t)+6e^{(2t)}\cos(3t)-9e^{(2t)}\sin(3t)$

Correct answer:

$a(t) =12e^{(2t)}\cos(3t)-5e^{(2t)}\sin(3t)$

Explanation:

The acceleration of the object can be found by differentiating the position equation twice. This can be done using the power rule, chain rule, and product rule.

The power rule is

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

The chain rule is

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

The product rule is

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

Using these three rules we can find the velocity equation from the position equation.

$v(t) = e^{(2t)}(2t^{(1-1)})\sin(3t) + e^{(2t)}\cos(3t)(3t^{(1-1)}) = 2e^{(2t)}\sin(3t) +3e^{(2t)}\cos(3t)$

Repeating this process gives us the acceleration

$\\a(t) = 2e^{(2t)}(2t^{(1-1)})\sin(3t) + 2e^{(2t)}\cos(3t)(3t^{(1-1)}) +\\ \\ 3e^{(2t)}(2t^{(1-1)})\cos(3t)+3e^{(2t)}(-\sin(3t))(3t^{(1-1)}) =\\ \\ 4e^{(2t)}\sin(3t)+12e^{(2t)}\cos(3t)-9e^{(2t)}\sin(3t)\\ \\ =12e^{2t}\cos(3t)-5e^{2t}\sin(3t)$

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Example Question #81 : How To Find Acceleration

If the acceleration function of an object is described by $a(t)=cos(2\pi)t$ , what is the acceleration when t=2 ?

Possible Answers:

Correct answer:

Explanation:

No deriving or integrating is necessary for this question.

Substitute the value of time into $a(t)=cos(2\pi)t$ .

$a(t=2)=cos(2\pi)(2)= 2cos(2\pi)= 2(1)=2$

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Example Question #471 : Calculus

Find the acceleration function of an object if the position function is x(t)= -ln(t)+cos(2t) .

Possible Answers:

$\frac{1}{t^2}-\frac{1}{4}cos(2t)$

$-\frac{1}{t^2}-4cos(2t)$

$-\frac{1}{t^2}+\frac{1}{4}cos(2t)$

$\frac{1}{t^2}-4cos(2t)$

$\frac{1}{t^2}+4cos(2t)$

Correct answer:

$\frac{1}{t^2}-4cos(2t)$

Explanation:

To find the acceleration function from the position curve, take the derivative twice.

$v(t)= \frac{d}{dx} x(t)$

x(t)= -ln(t)+cos(2t)

Recall the derivative of the natural log is a special case.

$f(x)=ln(u)\rightarrow f'(x)=\frac{1}{u}\cdot \frac{du}{dx}$

Therefore, the velocity functions becomes:

$v(t)=\frac{d}{dx}x(t)= -\frac{1}{t}-2sin(2t)= -t^-^1-2sin(2t)$

$a(t)=t^-^2-4cos(2t) =\frac{1}{t^2}-4cos(2t)$

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Example Question #81 : Acceleration

The veolcity of an object is given by the equation v(t) = 28t^3 + 10t^2 . What is the acceleration equation of the object?

Possible Answers:

$a(t) = 7t^4 + \frac{10t^3}{3}$

a(t) = 168t + 20

a(t) = 28t^4 + 10t^3

a(t) = 84t^2 + 20t

Correct answer:

a(t) = 84t^2 + 20t

Explanation:

The acceleration can be found by differentiating the velocity equation using the power rule where if

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

Using this rule we find the acceleration to be

$a(t) = 28(3)t^{(3-1)} + 10(2)t^{(2-1)} = 84t^2 + 20t$ .

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Example Question #82 : Acceleration

The position of an object is given by the function $x(t) = \sin(t)\cos(t)$ . What is the acceleration of the object at t =1 ?

Possible Answers:

$-2\sin(1)\cos^2(1) - 2\sin^2(1)\cos(1)$

$4\sin(1)\cos(1)$

$-4\sin(1)\cos(1)$

$-2\sin(1)\cos^2(1) + 2\sin^2(1)\cos(1)$

Correct answer:

$-4\sin(1)\cos(1)$

Explanation:

The acceleration of the object can be found by differentiating the position twice using the product rule where if

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

Applying this rule to the position gives us the velocity.

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

To differentiate a second time we must use the chain rule and the power rule where if

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

and if

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

Therefore the acceleration of the object is

$\\a(t) = (2)\cos^{2-1}(t)(-\sin(t)) - (2)\sin^{2-1}(t)(\cos(t)) \Rightarrow \\ \\ -2\cos(t)\sin(t) - 2\sin(t)\cos(t) = -4\sin(t)\cos(t)$ .

The acceleration at t =1 is then

$a(t) = -4\sin(1)\cos(1)$ .

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Example Question #83 : Acceleration

The position of an object is given by the equation x(t) = 10t^3 - t^4 . What is the acceleration of the object at t =1 ?

Possible Answers:

Correct answer:

Explanation:

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

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

Using this rule we find the velocity equation to be

$v(t) = 10(3)t^{(3-1)} - (4)t^{(4-1)} = 30t^2 - 4t^3$ .

Repeating this process gives the acceleration equation

$a(t) = 30(2)t^{(2-1)} - 4(3)t^{(3-1)} = 60t - 12t^2$ .

Therefore the acceleration at t =1 is

a(1) = 60(1) - 12(1)^2 = 60 - 12 = 48 .

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Example Question #84 : Acceleration

The position of an object is given by the equation x(t) = 20t^2 - 4t^3 + 5 . 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 position of the object twice using the power rule, where if

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

Applying this to the position equation gives us

$v(t) = 20(2)t^{2-1} - 4(3)t^{3-1} + 5(0)t^{(0-1)}= 40t - 12t^2$ .

Repeating this process gives us the acceleration equation

$a(t) = 40(1)t^{(1-1)} - 12(2)t^{(2-1)} = 40 - 24t$ .

Therefore the acceleration at t =2 is

a(2) = 40 - 24(2) = 40 -48 = -8 .

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Example Question #85 : Acceleration

The position of an object is given by the equation x(t) = 10t - 2t^2 - 5 . What is the acceleration of the object at t = 4 ?

Possible Answers:

Correct answer:

Explanation:

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

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

This rule gives a velocity equation of

$v(t) = 10(1)t^{(1-1)} - 2(2)t^{(2-1)} - 5(0)t^{(0-1)} = 10 - 4t$ .

Repeating this process give us an acceleration of

$a(t) = 10(0)t^{(0-1)} - 4(1)t^{(1-1)} = -4$ .

Therefore the acceleration of the object is constant at .

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Example Question #86 : Acceleration

The position of an object is given by the equation x(t) = 4t- 20 . What is the acceleration of the object at t =1 ?

Possible Answers:

Correct answer:

Explanation:

The acceleration of the object can be by differentiating the position equation using the power rule where if

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

Applying the power rule to the position equation gives

$v(t) = 4(1)t^{(1-1)} - 20(0)t^{(0-1)} = 4 - 0 = 4$ .

Repeating this process gives us the acceleration equation

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

Therefore the acceleration of the object is a constant .

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Example Question #85 : How To Find Acceleration

If an a ball was thrown across the field, where the path of the ball is described by a function of time s(t)=-2t^2+7t+4 , what is the acceleraton when the ball hits the ground?

Possible Answers:

Correct answer:

Explanation:

To know when the ball hits the ground, set the position s(t)=0 and solve for by factorization.

0=-2t^2+7t+4

0=-(2t^2-7t-4)

0=-(2t+1)(t-4)

0=(2t+1)(t-4)

2t+1=0

$t=-\frac{1}{2}$

t-4=0

t=4

Since time cannot be negative, the time when the ball hits the ground is t=4 .

To determine the acceleration of the ball, take the second derivative of the position function.

s(t)=-2t^2+7t+4

v(t)= -4t+7

a(t)=-4

The acceleration at t=4 is a(t)=-4 .

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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 #471 : Calculus

Example Question #81 : How To Find Acceleration

Example Question #471 : Calculus

Example Question #81 : Acceleration

Example Question #82 : Acceleration

Example Question #83 : Acceleration

Example Question #84 : Acceleration

Example Question #85 : Acceleration

Example Question #86 : Acceleration

Example Question #85 : How To Find Acceleration

All Calculus 1 Resources

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