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Calculus 1 : Acceleration

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Example Question #71 : Acceleration

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

Possible Answers:

a(t) = 0

None of the above.

$a(t) = 4\cos(t)\sin(t)$

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

Correct answer:

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

Explanation:

The acceleration of the object can be found by differentiating the position equation twice, using the product rule. The product rule is where

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

Using the product rule, the velocity of the object is

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

To differentiate the velocity equation we must use the power rule and the chain rule.

The power rule is

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

The chain rule is

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

Therefore the acceleration equation is then

$\\a(t) = v'(t) = (2)\cos^{(2-1)}(t)(-\sin(t)) - (2)\sin^{(2-1)}(t)(\cos(t)) \\ \Rightarrow -2\cos(t)\sin(t) - 2\sin(t)\cos(t) = -4\cos(t)\sin(t)$ .

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Example Question #72 : Acceleration

The velocity of an object is given by the equation v(t) = 2t + 5 . What is the acceleration of the object at time t = 4 ?

Possible Answers:

None of these.

Correct answer:

None of these.

Explanation:

The acceleration of the object is the derivative of the velocity. We can differentiate the velocity using the power rule where if

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

Therefore the acceleration equation of the object is

$a(t) = 2t^1 - 5t^0 =2(1)t^{(1-1)} - 5(0)t^{(0-1)} = 2$ .

Because the acceleration does not have depend on the variable the function is constant. Therefore the acceleration of the object is .

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Example Question #73 : Acceleration

The position of an object is given by the function x(t) = 5t^5 - 3t^2 . What is the acceleration of the object at time t =1 ?

Possible Answers:

Correct answer:

Explanation:

The acceleration of the object can be found by differentiating the position equation 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 equation is

$v(t) = 5(5)t^{(5-1)} - 3(2)t^{(2-1)} = 25t^4 - 6t^$ .

Repeating this process gives the acceleration equation

$a(t) = 25(4)t^{(4-1)} - 6(1)t^{(1-1)} = 100t^3 - 6$ .

Now we can find the acceleration at t= 1 .

a(1) = 100(1)^3 - 6 = 100 -6 = 94

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Example Question #74 : Acceleration

The jerk of an object is given by j(t) = 60t^3 . What is the acceleration of the object at time t= 2 , if the initial acceleration of the object is ?

Possible Answers:

200

249

Correct answer:

249

Explanation:

To find acceleration we integrate the jerk of the object. This can be done using the power rule where if

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

Therefore the acceleration is

$a(t) = \int (60t^3)dt = \frac{60t^{(3+1)}}{3+1} + C = 15t^4 +C$ .

The initial acceleration of the equation can now be used to find the value of .

a(0) = 15(0)^4 + C = 0 + C = 9

Therefore C = 9 and a(t) = 15t^4 + 9 .

The acceleration at time t =2 can now be found.

a(2) = 15(2)^4 + 9 = 249

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Example Question #75 : Acceleration

The velocity of an object is $v(t) = e^{7t}$ . What is the acceleration of the object at t = 0 ?

Possible Answers:

e^7

Correct answer:

Explanation:

The acceleration can be found by differentiating the velocity.

The velocity can be differentiated using the chain rule where

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

The acceleration is then

$a(t) = v'(t) = e^{7t}(7) = 7e^{7t}$ .

At t = 0 , the acceleration is

$a(0) = 7e^{7(0)} = 7e^0 = 7$ .

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Example Question #76 : Acceleration

The position of an object is given by the equation x(t) = 10t -17 . What is the acceleration equation for the object?

Possible Answers:

$a(t) = \frac{5t^3}{6} - \frac{17t^2}{2}$

a(t) = 10

None of these.

a(t) = 0

Correct answer:

a(t) = 0

Explanation:

The acceleration can be found by differentiating the position equation twice using the power rule.

The power rule is where

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

Therefore the velocity of the object is

$v(t) = x'(t) = 10(1)t^{(1-1)} - 17(0)t^{0-1} = 10$ .

The acceleration of the object can be found using the same method of differentiation as before.

$a(t) = v'(t) = 10(0)t^{(0-1)} = 0$

Therefore the acceleration of the object is .

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Example Question #77 : Acceleration

The velocity of an object is $v(t) = \sin(e^{2t})$ . What is the acceleration of the object at time t = -1 ?

Possible Answers:

$2e^2\cos(2)$

$\frac{2}{e^2}\cos(e^{-2})$

$\frac{2}{e^2}\cos(e^{2})$

${2e^2}\cos(e^{2})$

Correct answer:

$\frac{2}{e^2}\cos(e^{-2})$

Explanation:

The acceleration of the object is the derivative of the velocity equation. To differentiate this equation we can 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

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

From these rules we find that the acceleration of the object is

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

We can now find the acceleration at time t = -1 .

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

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Example Question #78 : Acceleration

The position of an object is given by the equation x(t) = 4t^2 - t^3 . 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 twice. This can be done using the power rule where if

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

Using this we can differentiate the velocity equation.

$v(t) = 4(2)t^{(2-1)} - (3)t^{(3-1)} = 8t - 3t^2$

Repeating this process we can find the acceleration equation.

$a(t) = 8(1)t^{(1-1)} - 3(2)t^{(2-1)} = 8 - 6t$

We can now find the acceleration at t = 2 .

a(2) = 8 - 6(2) = 8 - 12 = -4

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Example Question #79 : Acceleration

The position of an object is given by the equation x(t) = 6t^3e^t . What is the acceleration of the object?

Possible Answers:

a(t) = 36te^t+18t^2e^t+6t^3e^t

a(t) = 6te^t+36t^2e^t+6t^3e^t

a(t) = 36te^t+36t^2e^t+6t^3e^t

a(t) = 72te^t+36t^2e^t+6t^3e^t

Correct answer:

a(t) = 36te^t+36t^2e^t+6t^3e^t

Explanation:

To find the acceleration of the object we must differentiate the position of the object twice. We can differentiate the position using the power rule and the product rule where if

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

and if

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

Using these rules we obtian

$v(t) = (6(3)t^{(3-1)})e^t + 6t^3(e^t) = 18t^2e^t+6t^3e^t$ .

Using the same rules we can differentiate again to give the acceleration equation.

$a(t) = 18(2)t^{(2-1)}e^t +18t^2(e^t) + 6(3)t^{(3-1)}e^t + 6t^3(e^t) = 36te^t+36t^2e^t+6t^3e^t$

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Example Question #80 : Acceleration

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

Possible Answers:

a(t) = 6t

a(t) = 6

a(t) = 0

a(t) = -6

Correct answer:

a(t) = 6

Explanation:

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

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

Using this rule we find that

$v(t) = 3(2)t^{(2-1)} - 4(1)t^{(1-1)} = 6t - 4$ .

Repeating this process gives us

$a(t) = 6(1)t^{(1-1)} - 4(0)t^{(0-1)} = 6$ .

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Calculus 1 : Acceleration

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #71 : Acceleration

Example Question #72 : Acceleration

Example Question #73 : Acceleration

Example Question #74 : Acceleration

Example Question #75 : Acceleration

Example Question #76 : Acceleration

Example Question #77 : Acceleration

Example Question #78 : Acceleration

Example Question #79 : Acceleration

Example Question #80 : Acceleration

All Calculus 1 Resources

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