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Calculus 1 : How to find velocity

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Example Question #11 : Calculus

Find the velocity at $t=\pi$ if the position function of a spring is: $s(t)=5 \:cos\:t$ .

Possible Answers:

$\frac{1}{5}$

$-\frac{1}{5}$

Correct answer:

Explanation:

To find the velocity function, take the derivative of the position function.

$v(t)=\frac{\mathrm{d} }{\mathrm{d} t} s(t) = -5\:sin\:t$

Substitute $t=\pi$ into the velocity function.

$v(t=\pi)=-5\:sin\:\pi = 0$

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Example Question #1 : Derivatives

The position of an object is given by the following equation:

$s(t)=\frac{2}{3}t^3+\frac{5}{2}t^2-13t+27$

Determine the equation for the velocity of the object.

Possible Answers:

$v(t)=\frac{5}{2}t^2-13t+27$

v(t)=6t^2+2t-13

$v(t)=\frac{2}{3}t^3+\frac{5}{2}t^2-13t$

$v(t)=\frac{2}{3}t^2+\frac{5}{2}t+27$

v(t)=2t^2+5t-13

Correct answer:

v(t)=2t^2+5t-13

Explanation:

Velocity is the derivative of position, so in order to find the equation for the velocity of an object, all we must do is take the derivative of the equation for its position:

$s(t)=\frac{2}{3}t^3+\frac{5}{2}t^2-13t+27$

We will use the power rule to get the derivative.

$f(x)=ax^n \rightarrow f'(x)=a \cdot n x^{n-1}$

Therefore we get,

v(t)=s'(t)=2t^2+5t-13

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Example Question #12 : Calculus

The acceleration of an object is given by the following equation:

$a(t)=\frac{3}{4}t+\frac{5}{8}$

If $v(0)=1 \ m/s$ , find the velocity of the object at t=2 seconds.

Possible Answers:

$\frac{3}{2}$ m/s

$\frac{15}{4}$ m/s

m/s

$\frac{5}{8}$ m/s

Correct answer:

$\frac{15}{4}$ m/s

Explanation:

Acceleration is the derivative of velocity, so in order to obtain in equation for the velocity we must integrate the equation for acceleration with respect to time:

$v(t)=\int a(t)dt=\int \left(\frac{3}{4}t+\frac{5}{8}\right)dt=\frac{3}{8}t^2+\frac{5}{8}t+C$

Now we must use the given initial condition, v(0)=1, to solve for C:

$v(t)=\frac{3}{8}t^2+\frac{5}{8}t+C$

$v(0)=0+0+C=1\rightarrow C=1$

$v(t)=\frac{3}{8}t^2+\frac{5}{8}t+1$

Now we can simply plug in t=2 seconds to find the velocity of the object at that time:

$v(2)=\frac{3}{8}(2)^2+\frac{5}{8}(2)+1=\frac{12}{8}+\frac{10}{8}+1=\frac{30}{8}=\frac{15}{4}$ m/s

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Example Question #13 : Calculus

The displacement of a spacecraft is defined by y(x) = x^4+12x^2 , what is the instantaneous velocity of the ship at seconds?

Possible Answers:

142

180

190

204

Correct answer:

180

Explanation:

To find the velocity of the of the spacecraft, we can differentiate the position equation:

v(x) = 4x^3 + 24x

Now we can use x = 3 in the velocity equation to find the velocity at 3 seconds.

v(3) = 4(3)^3+24(3) = 108 + 72 = 180

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Example Question #2 : Derivatives

The position of a particle is represented by $f(x) = e^{x}+3x^2$ . What is the velocity at x=4 ?

Possible Answers:

Correct answer:

Explanation:

Differentiate the position equation, to get the velocity equation

$f(x) = e^{x}+3x^2$

$v(x) = e^x +3\cdot2 x^{2-1}=e^{x} + 6x$

Now we plug 4 into the equation to find the velocity

$v(4) = e^{(4)}+6(4)$

is approximately equal to 2.72. Therefore

$e^{(4)} + 6(4) = 2.72^{(4)} +24 \approx 55+24= 79$

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Example Question #14 : Calculus

What is the instantaneous velocity at $t = \frac{\pi}{2}$ of a particle whose positional equation is represented by $x(t) = 12\cos^2(t)$ ?

Possible Answers:

$-12\sqrt2$

Correct answer:

Explanation:

We find the velocity equation by differentiating the position equation. Since cos(t) is also a function we need to use the power rule along with the chain rule. This states to take the derivative of the outside function and multiply it by the derivative of the inside function. In math terms this is as follows:

$v(t) = 12\cdot 2 \cdot\cos(t)(-\sin(t)) = -24\cos(t)\sin(t)$

Using $\frac{\pi}{2}$ as the value of

$v\left(\frac{\pi}{2}\right) = -24\cos\left(\frac{\pi}{2}\right)\sin\left(\frac{\pi}{2}\right) = -24(0)(1) = 0$

Therefore the velocity at $t = \frac{\pi}{2}$ is 0

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Example Question #15 : Calculus

A weight hanging from a spring is stretched down by units from its rest position and released at time equals zero to bob up and down. Its position is given by x(t) = 5cos(t) . What is its velocity at time ?

Possible Answers:

$v(t) = 5\sin(t)$

$v(t) = -5\cos(t)$

v(t) = -5

$v(t) = -5\sin(t)$

Correct answer:

$v(t) = -5\sin(t)$

Explanation:

To find the velocity of the weight, we can differentiate the position equation.

Doing so gives us:

$v(t)=f(t)\cdot g'(t)+g(t)\cdot f'(t)$

where,

f(t)=5;f'(t)=0

$g(t)=\cos(t); g'(t)=-\sin(t)$

thus,

$v(t) = -5\sin(t)$

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Example Question #16 : Calculus

The position of a particle at time is given by s(t) = 4t^3-3t . What is the particle's velocity at time t= 7 ?

Possible Answers:

585

135

264

Correct answer:

585

Explanation:

To find the velocity, we must first find the velocity equation by differentiating the position equation of the particle.

v(t) = 12t^2-3

We can now use 7 as the value for to give

v(t) = 12(7)^2 - 3 = 585

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Example Question #11 : Spatial Calculus

The position function of a particle is given as: $s(t) = 9t^2+t-\sin(t)$ . What is the velocity function?

Possible Answers:

$v(t) = 18t+1-\cos(t)$

$v(t) = 18t+1+\cos(t)$

None of these answers are correct.

$v(t) = 18t-\cos(t)$

Correct answer:

$v(t) = 18t+1-\cos(t)$

Explanation:

To find the velocity function from the position function, we only need to differentiate the position function.

$v(t)=2\cdot 9 t^{2-1}+1t^{1-1}-\cos(t)$

$v(t) = 18t+1-\cos(t)$

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Example Question #12 : Spatial Calculus

Find the velocity of a function if the acceleration is: $a(t) = \sin(t) +2t$

Possible Answers:

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

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

$v(t) = \cos(t)+t^2$

$v(t) = -\cos(t)+\frac{t^2}{2}$

Correct answer:

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

Explanation:

To find the velocity, we integrate the acceleration equation

$v(t) = \int\sin(t)dt + \int2tdt$

$= -\cos(t)+\frac{2t^2}{2}$

$= -\cos(t)+t^2$

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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 #11 : Calculus

Example Question #1 : Derivatives

Example Question #12 : Calculus

Example Question #13 : Calculus

Example Question #2 : Derivatives

Example Question #14 : Calculus

Example Question #15 : Calculus

Example Question #16 : Calculus

Example Question #11 : Spatial Calculus

Example Question #12 : Spatial Calculus

All Calculus 1 Resources

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