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Calculus 1 : How to find position

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Example Questions

Example Question #61 : Position

Find the position function s(t) if v(t)=2t+3 and s(0)=4 .

Possible Answers:

s(t)=t^2+3t+4

s(t)=t^2+7

s(t)=t^2+3t

s(t)=t^2+7t

Correct answer:

s(t)=t^2+3t+4

Explanation:

In order to find the position function from the velocity function we need to take the integral of the velocity function.

$s(t)=\int v(t)dt$ .

When taking the integral, we will use the inverse power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ .

Applying this rule to each term we get,

$\\s(t)= \int v(t)=\int (2t+3)dt \\ \\s(t)=\frac{2t^2}{2}+3t+c\\ \\s(t)= t^2+3t+c$

To find the value of the constant c we will use the initial condition given in the problem.

Setting the initial condition t=0 ,

s(0)= 0^2+3(0)+c=4 yields c=4 .

Therefore the position function becomes,

s(t)= t^2+3t+4 .

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Example Question #61 : How To Find Position

Boulder question

A boulder rolls off the side of a seaside cliff that is above the water. The boulder is traveling at directly to the right when it leaves the cliff.

How far from the side of the cliff is the boulder when it lands in the water? Assume that gravity is the only force acting on the boulder, causing a downward acceleration of .

Possible Answers:

1673.3 m

1183.2 m

748.3m

3741.7m

529.2m

Correct answer:

1673.3 m

Explanation:

The vertical acceleration is given by a(t)=-9.8 .

Initial vertical velocity is .

To integrate we do,

$\int x^n=\frac{x^{n+1}}{n+1}$ .

Integrating once gives our velocity function, v(t)=-9.8t given our initial vertical velocity.

Integrating again gives a vertical position of y(t)= -4.9t^2+C .

Since we know the boulder starts at a height of 700 m , we determine that C=700 .

Solving 0=-4.9t^2+700 gives t=11.95 .

Then, determine a horizontal position equation for the boulder, with a starting position of and a constant velocity of 140m/s .

x(t)=140t

Evaluate x(t) when t=11.95 .

x(t)=140(11.95)=1673.3

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Example Question #63 : Position

A car is traveling at 12m/s when it is 80m along the highway. It accelerates constantly at a rate of 1.5m/s^2 .

Find the car's position as a function of .

Possible Answers:

x(t)=0.75t^2+12t

x(t)=0.75t^2+80

x(t)=0.75t^2+12t+80

x(t)=1.5t^2+12t+80

x(t)=3t^2+12t+80

Correct answer:

x(t)=0.75t^2+12t+80

Explanation:

The car has a constant of acceleration of 1.5m/s^2 .

Represent acceleration with a(t)=1.5 .

To integrate we do,

$\int x^n=\frac{x^{n+1}}{n+1}$ .

Integrate the acceleration function to get the velocity function.

v(t)=1.5t+C

Since initial velocity is 12m/s , you can solve for .

v(0)=12=1.5(0)+C

C=12

Thus, the function for velocity is v(t)=1.5t+12 .

Integrating the velocity equation gives the position function.

x(t)= 0.75t^2+12t+D

Since the initial position is 80m , you can solve for in the same way.

x(0)=80= 0.75(0)^2+12(0)+D

D=80

Finally, plugging back in gives the answer:

x(t)=0.75t^2+12t+80

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Example Question #64 : Position

Which of the following is perpendicular to the vector (10,2) ?

Possible Answers:

(10,-2)

(-10,-2)

(-2,11)

(2,-10)

(-10,2)

Correct answer:

(2,-10)

Explanation:

By definition, a given vector (a,b) has a perpendicular vector (b,-a) . Given a vector (10,2) , its perpendicular vector will be (2,-10) . We can further verify this result by noting that the product of two perpendicular vectors is ; since $(10,2)\times (2,-10)=(10\times 2)+(2\times -10)=20+(-20)=0$ , we know the two vectors are perpendicular to each other.

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Example Question #61 : How To Find Position

Which of the following is perpendicular to the vector (4,4) ?

Possible Answers:

(-2,4)

$\left(-\frac{1}{4},-\frac{1}{4}\right)$

(4,-4)

$\left(\frac{1}{4},\frac{1}{4}\right)$

(-4,-4)

Correct answer:

(4,-4)

Explanation:

By definition, a given vector (a,b) has a perpendicular vector (b,-a) . Given a vector (4,4) , its perpendicular vector will be (4,-4) . We can further verify this result by noting that the product of two perpendicular vectors is ; since $(4,4)\times (4,-4)=(4\times 4)+(4\times -4)=16+(-16)=0$ , we know the two vectors are perpendicular to each other.

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Example Question #65 : Position

The velocity of a particle is given by the function $v(t) = \frac{3}{8}t^2 + 6t - 9$ . What is the position of the particle at any time is the initial position of the particle s(0) = 8 ?

Possible Answers:

$s(t) = \frac{1}{8}t^3 + 9$

$s(t) = \frac{1}{8}t^3 + 3t^2 -9t + 8$

$s(t) = \frac{3}{4}t + 6$

$s(t) = \frac{3}{8}t^3 + 3t^2 + 8$

s(t) = 8t^3 - 12t^2 - 9t

Correct answer:

$s(t) = \frac{1}{8}t^3 + 3t^2 -9t + 8$

Explanation:

To find the position of particle given its velocity function,we must integrate it.

$s(t) = \int v(t) dt = \int (\frac{3}{8}t^2 + 6t - 9) dt = \frac{1}{8}t^3 + 3t^2 - 9t + C$

To solve for the constant of integration, we use the given initial position.

$s(0) = \frac{1}{8}(0)^3 + 3(0)^2 - 9(0) + C$

s(0)= 0 + 0 + C = 8

Therefore, the final equation is

$s(t) = \frac{1}{8}t^3 + 3t^2 -9t + 8$

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Example Question #8 : Integration

If the velocity of a particle is $\small v(t)=t^3-3t^2$ , and its position at $\small \small t=2$ is $\small \small \small s(2)=4$ , what is its position at $\small \small t=4$ ?

Possible Answers:

$\small \small s(4)=-4$

$\small s(4)=8$

$\small \small s(4)=0$

$\small \small s(4)=4$

Correct answer:

$\small s(4)=8$

Explanation:

This problem can be done using the fundamental theorem of calculus. We hvae

$\small \small \int_2^4 v(t)dt=\int_2^4 s'(t)dt=s(4)-s(2)$

where $\small s(t)$ is the position at time $\small t$ . So we have

$\small \small \small s(4)=s(2)+\int_2^4 v(t)dt$

so we need to solve for $\small \small \small \small \int_2^4 v(t)dt$ :

$\small \small \small \small \small \small \int_2^4 v(t)dt=\int_2^4 (t^3-3t^2)dt=\frac{t^4}{4}-t^3|_2^4$

$\small =\left(\frac{4^4}{4}-4^3\right)-\left(\frac{2^4}{4}-2^3\right)=(4^3-4^3)-\left(\frac{16}{4}-8\right)=-(4-8)=4$

So then the position at $\small t=4$ is

$\small \small s(4)=s(2)+\int_2^4 v(t)dt=4+4=8$

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Example Question #66 : Position

If the velocity of a particle is $\small v(t)=\sin2t$ and its initial position is $\small s(0)=3$ , what is its position at $\small t=\pi/2$ ?

Possible Answers:

$\small \small \small s(\pi/2)=3$

$\small \small s(\pi/2)=1$

$\small \small s(\pi/2)=4$

$\small \small \small s(\pi/2)=3.5$

Correct answer:

$\small \small s(\pi/2)=4$

Explanation:

We can use the fundamental theorem of calculus for this problem. We have

$\small \small \small \int_0^{\pi/2} v(t)dt =s(\pi/2)-s(0)$

so we have

$\small \small \small s(\pi/2)=s(0)+ \int_0^{\pi/2} v(t)dt$

This means we just need to solve $\small \int_0^{\pi/2} v(t)dt$ :

$\small \small \int_0^{\pi/2} v(t)dt=\int_0^{\pi/2} \sin 2tdt=-\frac{1}{2}\cos 2t|_0^{\pi/2}$

$\small \small =-\frac{1}{2}\left[ \cos \pi-cos 0 \right ]=-\frac{1}{2}[-1-1]=1$

So then the position of the particle at $\small t=\pi/2$ is

$\small \small \small \small s(\pi/2)=s(0)+ \int_0^{\pi/2} v(t)dt=3+1=4$

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Example Question #67 : Position

A car with an initial velocity of $10\frac{m}{s}$ begins to accelerate at a rate of $5\frac{m}{s^2}$ . How many meters will the car travel in the interval of time starting one second after it begins to accelerate and ending four seconds after it begins to accelerate?

Possible Answers:

67.5

30.5

37.5

Correct answer:

67.5

Explanation:

Begin by finding the velocity function, the integral of acceleration with respect to time:

$v(t)=\int a(t)dt =\int 5dt$

v(t)=5t + C

To find the constant of integration, the fact that the car has a velocity of $10 \frac{m}{s}$ when it begins to accelerate:

v(0)=10=5(0)+C

C=10

v(t)=5t+10

To find the distance travel, integrate the velocity function with respect to the lower and upper bounds, the start and end times respectively:

$d=\int_{t_i}^{t_f}v(t)dt=(2.5(t_f^2)+10(t_f))-(2.5(t_i^2)+10(t_i))$

d=(2.5(4^2)+10(4))-(2.5(1^2)+10(1))=(2.5(16)+40)-(2.5+10)

d=(80)-(12.5)=67.5

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Example Question #923 : Calculus

V(t) models the velocity of a spaceship. Find a function to model the spaceship's position as a function of time if P(t) contains the point (2,15) .

V(t)=8t^3-3t^2+5

Possible Answers:

P(t)=2t^4-t^3+5t-19

P(t)=2t^4-t^3+5

P(t)=24t^2-6t

P(t)=2t^4-t^3+5t

Correct answer:

P(t)=2t^4-t^3+5t-19

Explanation:

V(t) models the velocity of a spaceship. Find a function to model the spaceship's position as a function of time if P(t) contains the point (2,15).

Remember how position/velocity/acceleration are all related: Velocity is the first derivative of position, and acceleration is the derivative of velocity.

So to get to position, we need to integrate velocity.

$P(t)=\int V(t)dt=\int 8t^3-3t^2+5dt$

To integrate polynomials, we increase the exponent by 1, and then divide by the new exponent.

$\int 8t^3-3t^2+5dt=2t^4-t^3+5t+c$

Thus, our function so far looks like:

2t^4-t^3+5t+c

However, we know that it must pass through the point (2,15), so we have to plug the point in and solve for C

15=2(2)^4-(2)^3+5(2)+c

15=32-8+10+c

c=-19

So our answer is:

2t^4-t^3+5t-19

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Calculus 1 : How to find position

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #61 : Position

Example Question #61 : How To Find Position

Example Question #63 : Position

Example Question #64 : Position

Example Question #61 : How To Find Position

Example Question #65 : Position

Example Question #8 : Integration

Example Question #66 : Position

Example Question #67 : Position

Example Question #923 : Calculus

All Calculus 1 Resources

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