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Calculus 1 : Velocity

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Example Questions

Example Question #51 : Calculus

The position of an object is given by the equation $x(t) = \cos^2(t)$ . What is the velocity of the object when t =0 ?

Possible Answers:

$\frac{\sqrt2}{3}$

$\frac{1}{2}$

Correct answer:

Explanation:

Given the equation $x(t) = \cos^2(t)$ , the velocity can be found by differentiating the position equation. To properly differentiate this equation, we must use the chain rule where if

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

Therefore, the derivative of the function is,

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

We now use t = 0 to find the velocity.

$v(0) = -2\cos(0)\sin(0) = -2(1)(0) = 0$

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Example Question #52 : Spatial Calculus

The acceleration of an object is given by the equation a(t) =40 . What is the velocity at t = 3 if the object has an initial velocity of ?

Possible Answers:

140

120

100

Correct answer:

140

Explanation:

The velocity of the object is the integral of the accelertion. We can use the power rule to integrate the equation. The power rule is that

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

Therefore

$v(t) = \int (40)dt = \int (40t^0)dt = \frac{40t^{(0+1)}}{0+1} +C = 40t^1 +C$

To find the velocity we must now solve for the value of , using the initial velocity of the object.

v(0) = 40(0) +C = 0 + C= 20

Therefore C = 20 and v(t) = 40t + 20

We can now find the velocity at time t = 3

v(3) = 40(3) +20 = 120 +20 = 140

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Example Question #51 : How To Find Velocity

The position of an object is given by the equation $x(t) = \sin^2(t) + \cos^2(t)$ . What is the velocity equation of the object?

Possible Answers:

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

v(t) = 0

$v(t) = -\sin(t)\cos(t)$

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

Correct answer:

v(t) = 0

Explanation:

The velocity of the object can be found by differentiating the position equation. To differentiate the position equation 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 where

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

Therefore the velocity equation is

$v(t) = x'(t) = 2\sin(t)\cos(t) + 2\cos(t)(-\sin(t)) \Rightarrow \\ \Rightarrow 2\sin(t)\cos(t) - 2\sin(t)\cos(t) = 0$

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Example Question #1 : Calculus Review

The position of an object is given by the equation x(t) = e^t +t . What is the velocity of the object at t = 2 ?

Possible Answers:

15.5

8.3

9.3

Correct answer:

8.3

Explanation:

The velocity of the object can be found by differentiating the position equation of the object. To differentiate the position equation of the object, we can use the power rule for the second term where if

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

Using this rule we find that

$v(t) = x'(t) = e^t + (1)t^{(1-1)} = e^t +1$

We can now use the value of t = 2 to solve for the velocity at t = 2

$v(2) = e^2 +1 \approx 2.7^2 +1 = 8.3$

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Example Question #51 : Spatial Calculus

The position of an object is given by the equation $x(t) = 7e^t - \cos^2(t)$ . What is the equation for the velocity of the object?

Possible Answers:

$v(t) = 7e^t - \sin(t)\cos(t)$

$v(t) = 7te^t + \sin(t)\cos(t)$

$v(t) = 7 - 2\sin(t)\cos(t)$

$v(t) = 7e^t + 2\sin(t)\cos(t)$

Correct answer:

$v(t) = 7e^t + 2\sin(t)\cos(t)$

Explanation:

The velocity of the object is the derivative of the position equation. To differentiate the position equation 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

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

Applying these two rules to the position equation gives us

$v(t) = x'(t) = 7e^t - 2\cos(t)(-\sin(t)) = 7e^t +2\sin(t)\cos(t)$

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Example Question #51 : How To Find Velocity

The position of a car is defined as $p(t)=5t^{2}-3t+7$ . What is the car's velocity at t=3 ?

Possible Answers:

Correct answer:

Explanation:

The velocity v(t) is defined as the first derivative of its position p(t) , or v(t)=p'(t) .

Using the power rule to differentiate,

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

We find.

$p(t)=5t^{2}-3t+7$ , v(t)=p'(t)=10t-3 .

Swapping t=3 , v(t)=10(3)-3=30-3=27 .

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Example Question #57 : Spatial Calculus

The position of a particle is defined as $p(t)=7t^{2}+4t+3$ . What is the particle's velocity at t=5 ?

Possible Answers:

Correct answer:

Explanation:

The velocity v(t) is defined as the first derivative of its position p(t) , or v(t)=p'(t) .

Using the power rule when differentiating,

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

we find,

$p(t)=7t^{2}+4t+3$ , v(t)=p'(t)=14t+4 .

Swapping t=5 , v(t)=14(5)+4=70+4=74 .

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Example Question #58 : Spatial Calculus

The position of a particle is defined as $p(t)=3t^{2}+2t+10$ . What is the particle's velocity at t=1 ?

Possible Answers:

Correct answer:

Explanation:

The velocity v(t) is defined as the first derivative of its position p(t) , or v(t)=p'(t) .

Use the power rule to differentiate the position function.

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

Since $p(t)=3t^{2}+2t+10$ , v(t)=p'(t)=6t+2 .

Swapping t=1 , v(t)=6(1)+2=6+2=8 .

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Example Question #51 : How To Find Velocity

The acceleration of an object is given by the equation a(t) = 12t^2 - 4t^3 . What is the velocity of the object at time t = 1 , if the velocity of the object is at time t = 3 ?

Possible Answers:

Correct answer:

Explanation:

The velocity of the object can be found by integrating the acceleration of the object. This can be done using the power rule where if

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

Integrating the acceleration equation gives us

$v(t) = \int (12t^2 - 4t^3)dt = \frac{12t^{(2+1)}}{2+1} - \frac{4t^{(3+1)}}{3+1} + C = 4t^3 - t^4 +C$

We can solve for the value of using the velocity of the object at t = 3

v(3) = 4(3)^3 - (3)^4 + C =108 - 81+C = 27+C = 3

Therefore C= -24 and v(t) = 4t^3 - t^4 -24

We can now find the velocity at time t =1

v(1) = 4(1)^3 - (1)^4 -24 = 4 - 1 -24 = -21

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Example Question #52 : How To Find Velocity

The position of an object is given by the equation $x(t) = 13e^{3t} - 3$ . What is the velocity of the object?

Possible Answers:

$v(t) = 3e^{3t} - 3t$

$v(t) = 13e^{3t} - 3t$

$v(t) = 39e^{3t}$

$v(t) = 13e^{3t}$

Correct answer:

$v(t) = 39e^{3t}$

Explanation:

The velocity of the object can be found by differentiating the position equation of the object. To accurately 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 where if

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

Therefore the velocity of the object is

$v(t) = 13e^{3t}(3(1)t^{1-1}) - 3(0)t^{(0-1)} = 39e^{3t} - 0 = 39e^{3t}$

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Calculus 1 : Velocity

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #51 : Calculus

Example Question #52 : Spatial Calculus

Example Question #51 : How To Find Velocity

Example Question #1 : Calculus Review

Example Question #51 : Spatial Calculus

Example Question #51 : How To Find Velocity

Example Question #57 : Spatial Calculus

Example Question #58 : Spatial Calculus

Example Question #51 : How To Find Velocity

Example Question #52 : How To Find Velocity

All Calculus 1 Resources

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