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Calculus 1 : How to find velocity

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Example Question #81 : Calculus

The position of an object is given by the function $x(t) = \sin(6t^3)$ . What is the velocity of the object at t= 0 ?

Possible Answers:

Correct answer:

Explanation:

To find the velocity of the object we can differentiate the position of the object. 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

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

Therefore at t =0 the velocity of the object is

$v(0) = 18(0)^2\cos(6(0)^3) = (0)\cos(0) = 0$ .

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Example Question #82 : Calculus

The position of an object is given by the equation $\sin(t)\cos^2(t)$ . What is the velocity of the object at t = 1 ?

Possible Answers:

$-\cos^3(1) + 2\sin(1)\cos^2(1)$

$\cos^3(1) + 2\sin(1)\cos^2(1)$

$\cos^3(1) + 2\sin^2(1)\cos(1)$

$\cos^3(1) - 2\cos(1)\sin^2(1)$

Correct answer:

$\cos^3(1) - 2\cos(1)\sin^2(1)$

Explanation:

The velocity of the object can be found by differentiating the position. This can be done using the power rule, the chain rule, and the 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

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

Appying these rules to the position equation gives us

$v(t) = (\cos(t))\cos^2(t) + \sin(t)(2\cos(t)(-\sin(t))) = \cos^3(t) - 2\cos(t)\sin^2(t)$ .

Using the velocity equation we find the velocity at t =1 to be

$v(1) = \cos^3(1) - 2\cos(1)\sin^2(1)$

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Example Question #83 : Calculus

The position of an object is given by the equation $x(t) = -\cos(3t)$ . What is the velocity of the function at time t = 1 ?

Possible Answers:

$3\sin(1)$

$3\sin(3)$

$\sin(3)$

Correct answer:

$3\sin(3)$

Explanation:

The velocity if the function can be found by differentiating the position equation. This can be done using the power rule and the chain rule where if

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

and if

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

Using these rules we obtain

$v(t) = -(-\sin(3t))(3(1)t^{(1-1)}) = 3\sin(3t)$ .

We can now find the velocity at t = 1 .

$v(1) = 3\sin(3(1)) = 3\sin(3)$

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Example Question #84 : Calculus

The acceleration of the an object is given by the equation a(t) = 10t^4 + 6t^5 - 6t . What is the velocity of the object if the object has an initial velocity of ?

Possible Answers:

v(t) = 2t^5 + t^6 + 3t^2

v(t) = 2t^5 + 3t^6 + 3t^2

v(t) = 2t^5 + t^6 - 3t^2

v(t) = 2t^5 + t^5 - 3t^2

Correct answer:

v(t) = 2t^5 + t^6 - 3t^2

Explanation:

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

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

Using this rule we get the equation

$v(t) = \int (10t^4 + 6t^5 - 6t)dt = \frac{10t^{(4+1)}}{4+1} + \frac{6t^{(5+1)}}{5+1} - \frac{6t^{(1+1)}}{1+1} +C$

v(t) = 2t^5 + t^6 - 3t^2 +C .

Using the initial velocity of the object we can find the value of .

v(0) = 2(0)^5 + (0)^6 - 3(0)^2 +C = 0 + 0 -0 +C = 0

Therefore C = 0 and v(t) = 2t^5 + t^6 - 3t^2 .

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Example Question #85 : Calculus

Below is the graph of an object's position over time. Place the following in order from largest to smallest:

The instantaneous velocity at t=0

II. The average velocity between t=0.8 and t=2.2

III. The average velocity between t=3.12 and t=4.04

IV. The instantaneous velocity at t=2.2

Graph4

Possible Answers:

I, IV, III, II

I, II, IV, III

I, III, IV, II

II, IV, I, III

I, IV, II, III

Correct answer:

I, II, IV, III

Explanation:

The average velocity of an object over a given interval is given by $\frac{\bigtriangleup position }{\bigtriangleup time}$ and is represented on a position graphs by the slope of the secant line. The instantaneous velocity at a given time is represented by the slope of the tangent line at a point. Notice that the velocity is positive and decreasing. It is clear that is largest and is greater than III . Additionally, since the velocity is decreasing, is greater than . Hence we have: I, II, IV, III .

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Example Question #86 : Calculus

A football is launched into the air during kick off. It's position is given by the function below. Find its instantaneous velocity at the time when it is caught by the kick returner. Assume the kick returner catches the ball at chest-level (about 4 feet off of the ground.)

p(t)=-16t^2+64t+2

Possible Answers:

v=-63ft/s

v=63ft/s

v=0ft/s

v=-127ft/s

Correct answer:

v=-63ft/s

Explanation:

The question asks us to find the slope of the tangent line at the time when the ball is caught. We can either use the limit definition of the derivative to find the slope directly or take the derivative, then plug in the appropriate time value to get the slope. I will be using the second method as it is a bit simpler.

We begin by solving for the time at which the ball is caught.

$-16t^2+64t+2=4 \implies t=.031s,3.97s$

Unless new rules come out for onsides kicks, we can choose the larger time value. Then we take the derivative:

p'(t)=-32t+64

We substitute the appropriate time value in to get:

p'(3.97)=-63ft/s

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Example Question #87 : Calculus

Given the velocity function v=3t^2+2 , what is the average velocity from t=1 to t=2 ?

Possible Answers:

Correct answer:

Explanation:

Write the formula for average velocity.

$V_{avg}=\frac{X_f-X_i}{t_f-t_i}$

Integrate the velocity function to determine the position function.

$x(t)=\int v dt=\int (3t^2+2) \:dt$

x(t)= t^3+2t

Substitute the value of t=1 to determine X_i , or initial position.

x(1)= 1^3+2(1) =1+2=3

Substitute the value of t=2 to determine X_f , or final position.

x(2)= 2^3+2(2) =8+4=12

Substitute all the known values into the average velocity formula.

$V_{avg}=\frac{X_f-X_i}{t_f-t_i}= \frac{12-3}{2-1}=9$

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Example Question #88 : Calculus

The position of an object is given by the function $x(t) = \sin(t)\cos(t)$ . What is the velocity of the object at t =3 ?

Possible Answers:

$\frac{\cos^2(3) - \sin^2(3)}{\cos^2(3)}$

None of these.

$\cos^2(3) + \sin^2(3)$

$\cos^2(3) - \sin^2(3)$

Correct answer:

$\cos^2(3) - \sin^2(3)$

Explanation:

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

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

Therefore the velocity of the object is

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

We can now solve for the velocity of the object at t =3 .

$v(3) = \cos^2(3) - \sin^2(3)$

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Example Question #89 : Calculus

The position of an object is given by the equation x(t) = 10 . What is the velocity of the object at t =3 ?

Possible Answers:

Correct answer:

Explanation:

The velocity of the object can be found by differentiating the position. The differentiation can be done using the power rule where if

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

Then the velocity of the object is

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

Therefore the velocity of the object is a constant .

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Example Question #90 : Calculus

The acceleration of an object is given by the equation a(t) =8t^3 + 12t . What is the velocity of the equation at t =1 , if the velocity is at t =3 ?

Possible Answers:

Correct answer:

Explanation:

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

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

Using this rule we find that

$v(t) = \int (8t^3 + 12t)dt = \frac{8t^{(3+1)}}{3+1} + \frac{12t^{1+1}}{1+1} + C = 2t^4 + 6t^2 +C$ .

We can find the value of using the velocity at t =3

v(3) = 2(3)^4 + 6(3)^2 +C = 162 + 54 +C = 216 +C = 0 .

Therefore C = -216 and v(t) = 2(t)^4 + 6t^2 - 216 .

We can now solve for the velocity of the object at t =1 .

v(1) = 2(1)^4 +6(1)^2 - 216 = 2 + 6 - 216 = -208

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Calculus 1 : How to find velocity

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #81 : Calculus

Example Question #82 : Calculus

Example Question #83 : Calculus

Example Question #84 : Calculus

Example Question #85 : Calculus

Example Question #86 : Calculus

Example Question #87 : Calculus

Example Question #88 : Calculus

Example Question #89 : Calculus

Example Question #90 : Calculus

All Calculus 1 Resources

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