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Calculus 1 : Velocity

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Example Questions

Example Question #181 : Calculus

Find the velocity of an bullet after seconds if its inital velocity is 1200 ft/s and its acceleration is ft/s.

Possible Answers:

v=976

v=1200

v=-32

v=2

v=500

Correct answer:

v=976

Explanation:

To find the velocity of the object after a certain time, we first must find the velocity function of the object, then solve the position function at that given time.

The velocity function of an object moving with a uniform, or constant, acceleration is $v(t)=v_{0}+at$ , where $v_{0}$ is the inital velocity of the object and is the acceleration of the object.

For this problem:

$v_{0}=1200ft/s$ , $a=-32ft/s^{2}$

v(t)=1200-32t

v(7)=1200-32(7)=1200-224=976

After seconds, the velocity of bullet is 976 ft/s.

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Example Question #182 : Calculus

Find the velocity function of an bullet after it has been fired if its inital velocity is 1200 ft/s and its acceleration is ft/s.

Possible Answers:

v(t)=1200-32t

v(t)=5

v(t)=1200t-32

v(t)=1200

v(t)=-32t

Correct answer:

v(t)=1200-32t

Explanation:

The velocity function of an object moving with a uniform, or constant, acceleration is $v(t)=v_{0}+at$ , where $v_{0}$ is the inital velocity of the object and is the acceleration of the object.

For this problem:

$v_{0}=1200ft/s$ , $a=-32ft/s^{2}$

v(t)=1200-32t

Side note: The velocity function is the first derivative of the position function moving with a uniform acceleration $s(t)=s_{0}+v_{0}t+\frac{1}{2}at^{2}$

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Example Question #183 : Calculus

If the position of a particle is given by the function $x(t)=t+t^2e^{-t^2}$ , what is its velocity at time t=2 ?

Possible Answers:

$2-12e^{-4}$

$1-12e^{-4}$

1+12e^4

$-12e^{-4}$

$2+12e^{-4}$

Correct answer:

$1-12e^{-4}$

Explanation:

Velocity is the time derivative of position:

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

For the position function

$x(t)=t+t^2e^{-t^2}$

Velocity is thus:

$v(t)=1+2te^{-t^2}-2t^3e^{-t^2}$

$v(2)=1+4e^{-4}-16e^{-4}$

$v(2)=1-12e^{-4}$

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Example Question #181 : Calculus

The position of an animal is given by the function

$x(t)=\frac{1}{3}t^3-3t^2+12t$ .

At what time is the animal's velocity at its lowest?

Possible Answers:

t=9

t=10

t=3

t=0

t=6

Correct answer:

t=3

Explanation:

Differentiating the position function will result in the function representing the animal's velocity.

$x(t)=\frac{1}{3}t^3-3t^2+12t$

v(t)=t^2-6t+12

Complete the square to rearrange the function as follows:

v(t)=t^2-6t+12-3+3

v(t)=t^2-6t+9+3

v(t)=(t-3)^2+3

Notice that this function is an upward facing parabola with its vertex at (3, 3) .

Since the parabola is upward facing (concave up), the vertex will be a local minimum.

Therefore, the velocity is lowest at the vertex, when t=3 .

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Example Question #184 : Calculus

The position of a particle is modeled by the function $x(t)=t^2-\frac{1}{3}t^3$ . At what time, , will the particle reverse directions?

Possible Answers:

1.5

Correct answer:

Explanation:

Velocity is the time derivative of position:

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

For the function

$x(t)=t^2-\frac{1}{3}t^3$

Velocity is thus:

v(t)=2t-t^2

There will be a change in direction of the particle after the velocity reaches zero:

v(t_f)=0=2t_f-t_f^2

t_f(2-t_f)=0

Neglecting the trivial solution of t_f=0 , the particle will turn around at t_f=2 .

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Example Question #186 : Calculus

Find the velocity function of a particle if its position function is s(t)=sin(2t)-cos(3t) .

Possible Answers:

v(t)=2cos(2t)+3sin(3t)

$v(t)=2\pi$

v(t)=sin(2t)-cos(3t)

v(t)=sin(2t)

v(t)=-cos(3t)

Correct answer:

v(t)=2cos(2t)+3sin(3t)

Explanation:

The velocity function is the first derivative of the position function, or

$v(t)=s'(t)=\frac{d}{dt}(s(t))$ .

We will use the trigonometric rules,

$\frac{d}{dt}(cos(at))=-asin(at)$ and

$\frac{d}{dt}(sin(at))=acos(at)$ ,

where is a constant, to find the derivative of this position function.

For this problem:

s(t)=sin(2t)-cos(3t)

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

v(t)=2cos(2t)+3sin(3t)

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Example Question #187 : Calculus

Find the velocity function of the particle given the following position function:

$s(t)=x^2+2x+\csc(x)$

Possible Answers:

$2x+2-\csc(x)\cot(x)$

$x^2+2x-\csc(x)\cot(x)$

$4x-\csc(x)\cot(x)$

$2x+2+\csc(x)\cot(x)$

Correct answer:

$2x+2-\csc(x)\cot(x)$

Explanation:

The velocity function is the derivative of the position function:

$v(t)=s'(t)=2x+2-\csc(x)\cot(x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

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Example Question #185 : Calculus

x(t)=ln(t)

The function x(t) represents the position of a particle. What will the particle's velocity when t=4 ?

Possible Answers:

0.10

0.25

Not enough information. The particle's acceleration function must be known.

Correct answer:

0.25

Explanation:

To find the function representing velocity, differentiate the position function.

The first derivative is taken as follows:

x(t)=ln(t)

$x'(t)=\frac{1}{t}$

Therefore, the velocity function is:

$v(t)=\frac{1}{t}$

Next, evaluate the velocity function when t=4 .

In other words find v(4) .

$v(4)=\frac{1}{4}=0.25$

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Example Question #186 : Calculus

When t=0 , a pendulum is swinging at a velocity of 4 m/s . It accelerates according to the following function:

a(t)= 2cos(2t)

Find the function representing the pendulum's velocity.

Possible Answers:

v(t)=sin(2t)+4

$v(t)=e^{(2t)+4}{}$

$v(t)=\frac{1}{cos(2t)}$

v(t)=tan(t)+2

v(t)=2cos(2t)+4

Correct answer:

v(t)=sin(2t)+4

Explanation:

To find the function representing velocity, you must find the indefinite integral of the acceleration function.

Antidifferentiating the acceleration function is done as follows:

a(t)= 2cos(2t)

v(t)=sin(2t)+C

From the problem, you know that when t=0 , the pendulum has a velocity of 4 m/s . Therefore, v(0)=4 .

Substitute these values in:

4=sin(2*0)+C

4=0+C

4=C

Therefore, your velocity function is:

v(t)=sin(2t)+4

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Example Question #190 : Calculus

A car is moving with a position function

$s(t)=e^{2t}+3t^2$

What is the velocity of the car at t=0 ?

Possible Answers:

Correct answer:

Explanation:

The velocity function is the first derivative of the position function:

$v(t)=s'(t)=2e^{2t}+6t$

and was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Now, plug in the time t=0 into the velocity function:

v(0)=2

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Calculus 1 : Velocity

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #181 : Calculus

Example Question #182 : Calculus

Example Question #183 : Calculus

Example Question #181 : Calculus

Example Question #184 : Calculus

Example Question #186 : Calculus

Example Question #187 : Calculus

Example Question #185 : Calculus

Example Question #186 : Calculus

Example Question #190 : Calculus

All Calculus 1 Resources

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