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Calculus 1 : Calculus

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Example Questions

Example Question #161 : How To Find Velocity

Find the velocity function for a particle given its poision as a function of time.

p(t)=15t^4+5t^3-t+20

Possible Answers:

v(t)=15t^3+15t^2-1

v(t)=60t^3+15t^2

v(t)=60t^3+15t^2-1

v(t)=20t^3+15t^2-1

None of these

Correct answer:

v(t)=60t^3+15t^2-1

Explanation:

Velocity is the derivative of position. The derivative of $x^{n}$ is $nx^{n-1}$ .

With this information the velocity function can be found.

v(t)=60t^3+15t^2-1

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Example Question #161 : How To Find Velocity

The acceleration function of a moving object is a(t)=4.9 and the intial position of said object is zero. If the object has traveled 500 meters in seconds, find the velocity of the object at seconds.

Possible Answers:

$51\frac{m}{s}$

None of the above.

$50\frac{m}{s}$

$49\frac{m}{s}$

$48\frac{m}{s}$

Correct answer:

$50\frac{m}{s}$

Explanation:

In order to solve this problem, you must know that the derivative of a position function is the veloctiy function and the derivative fo the velocity function is the acceleration function. Thefore, by taking the double integral of the acceleration function we will be able to find the position function.

In order to take the integral of the acceleration function, a(t)=4.9 , we must first know the general rule of integration $\int x^n dx=\frac{x^{n+1}}{n+1}+C$ .

Taking the integral of the acceleration function, we find that the veloctiy function is equivalent to v(t)=4.9t+C . We must somehow figure out how to solve for .

Taking the integral of the veloctiy function to get the position function, we find that the position function is S(t)=2.45t^2+Ct . Note that this time another isn't included because the intial position of the obejct is 0.

We know that the object traveled 500 meters in 10 seconds, therefore by plugging in S(10)=500 we can solve for .

500=2.45(10)^2+10C

C=25.5

Therefore the veloctiy function becomes v(t)=4.9t+25.5 . The problem asks us to find the veloctiy of the object at 5 seconds, plugging in 5 we find that the velocity of the object is $50\frac{m}{s}$ .

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Example Question #162 : Calculus

Find the velocity function v(t) if the position function is s(t)=t^2 -6t +5 .

Possible Answers:

v(t)=2t^2-1

v(t)=t^2-6t

v(t)=2t+5

v(t)=2t-6

v(t)=t^2-t

Correct answer:

v(t)=2t-6

Explanation:

Velocity is equal to the derivative of position with respect to time.

The derivative of this position function, by the power rule which states,

$\frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}$ is

$\\s(t)=t^2 -6t +5 \\s'(t)=2t^{2-1}-6t^{1-1}$

v(t)=s'(t) = 2t - 6 .

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Example Question #161 : Spatial Calculus

Find the average velocity on the interval $0 \leq t \leq 5$ for s(t) = t^2 - 4t+15 .

Possible Answers:

Correct answer:

Explanation:

Average velocity is defined as $v_{avg} = \frac{\Delta s}{\Delta t}$ .

Therefore, on the interval $0 \leq t \leq 5$ ,

$\\v_{avg} = \frac{s(5)-s(0)}{5-0}\\ v_{avg} = \frac{20 - 15}{5}\\ v_{avg} = 1$

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Example Question #161 : Velocity

Find the instantaneous velocity at t = 4 for the position function $s(t) = 3ln(t) + \frac{t^3}{16}$ .

Possible Answers:

4.75

3.25

1.75

3.75

4.25

Correct answer:

3.75

Explanation:

The instantaneous velocity can be found by taking the derivative of the position function.

Recall the following rules of differentiation to help solve this problem.

Power Rule:

$\frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}$

Natural Log Rule:

$\frac{\mathrm{d} }{\mathrm{d} t}(aln(bt)) = \frac{a}{t}$

The instantaneous velocity at t=4 , by the power rule and the rule for natural logs, is

$\\v(4) = s'(4) \\ v(t) = s'(t) = \frac{3}{t}+\frac{3t^2}{16}\\ v(4) = \frac{3}{4} + 3\\ \\v(4) = 3.75$

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Example Question #165 : Calculus

Find the velocity function v(t) if the position function is $s(t) = \frac{3}{t}+ \sqrt{t}$ .

Possible Answers:

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

$v(t) = -\frac{3}{t} + \frac{1}{2\sqrt{t}}$

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

$v(t) = -\frac{3}{t^2} + \frac{1}{2\sqrt{t}}$

$v(t) = -\frac{3}{t^2} + \frac{1}{\sqrt{t}}$

Correct answer:

$v(t) = -\frac{3}{t^2} + \frac{1}{2\sqrt{t}}$

Explanation:

Velocity is the derivative of position with respect to time.

Recall the following rule of differentiation to help solve this problem.

Power Rule:

$\frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}$

Therefore, by the power rule, remembering that $\frac{3}{t} = 3t^{-1}$ and $\sqrt{t} = t^{1/2}$ , the velocity function is

$s'(t) = v(t) = -\frac{3}{t^2} + \frac{1}{2\sqrt{t}}$ .

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Example Question #161 : How To Find Velocity

Find the velocity function v(t) if the position function is s(t) = tcos(t) .

Possible Answers:

v(t) = cos(t) - tsin(t)

v(t) = tcos(t) - sin(t)

v(t) = tcos(t) - tsin(t)

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

v(t) = cos(t) + tsin(t)

Correct answer:

v(t) = cos(t) - tsin(t)

Explanation:

Velocity is equal to the derivative of position with respect to time.

Recall the following rules of differentiation to help solve this problem.

Power Rule:

$\frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}$

Differentiation rules for sine and cosine:

$\\\frac{\mathrm{d} }{\mathrm{d} t}(asin(bt)) = abcos(bt)\\ \frac{\mathrm{d} }{\mathrm{d} t}(acos(bt)) = -absin(bt)\\$

Product rule:

$\frac{\mathrm{d} }{\mathrm{d} t}(u(t)v(t)) = u(t)v'(t)+u'(t)v(t)$

Since s(t) = tcos(t) , then, by the product rule, the power rule, and the differentiation rules for sine and cosine,

v(t) = cos(t) - tsin(t) .

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Example Question #163 : How To Find Velocity

Given the position function find the velocity function

p(t)=4x^3-5x^2-x+15 .

Possible Answers:

12x^2-10x-1

x^2-10x-1

12x^2-10x+14

None of these

12x^3-10x^2-x

Correct answer:

12x^2-10x-1

Explanation:

The velocity is just the derivative of the position.

The power rule states that the derivative of x^n is $nx^{n-1}$ .

This is the only rule we need to find the derivative of the position function.

v(t)=12x^2-10x-1

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Example Question #164 : How To Find Velocity

Find the average velocity of the function over the given interval.

Given the velocity function,

g(t)=5+tan(t) , on the interval $\left [ -\frac{\pi }{4},\frac{\pi }{4} \right ]$ .

Possible Answers:

$\frac{4\pi }{3}$

$2\pi$

$\frac{21\pi }{6}$

$\pi$

Correct answer:

Explanation:

Here we need the average velocity.

An average velocity can be found by taking the integral of the velocity function then taking the difference of that function at the given bounds and dividing that by the interval. In other words:

$V_{avg}=\frac{1}{b-a}\int_a^b g(t)\ dt$

We are given the velocity function, so putting the pieces together we get:

$V_{avg}=\frac{1}{\frac{\pi}{4}--\frac{\pi}{4}}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 5+tan(t)$

$V_{avg}=\frac{1}{\frac{\pi}{2}}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 5+tan(t)$

$V_{avg}={\frac{2}{\pi}}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 5+tan(t)$

We will use the power rule to integrate the constant which states,

$x^n\rightarrow \frac{x^{n+1}}{n+1}$ .

To integrate tangent we will use the rules of trigonometric functions which states that,

$\int tan(x)=-ln|cos(x)|$ .

Applying these rules we get the following:

$V_{avg}=\frac{2}{\pi}\left[5t+-ln|cos(t)|\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}$

Now plugging in our initial and final values we get our answer,

$V_{avg}=\frac{2}{\pi}\left[\left[5\left(\frac{\pi}{4} \right )+-ln\left|cos\left(\frac{\pi}{4} \right )\right|\right]-\left[5\left(\frac{-\pi}{4} \right )+-ln\left|cos\left(\frac{-\pi}{4} \right )\right|\right]\right]$

$V_{avg}=\frac{2}{\pi}\left[\left[\left(\frac{5\pi}{4} \right )+-ln\left|\frac{1}{\sqrt2} \right|\right]-\left[\left(\frac{-5\pi}{4}\right)+-ln\left|\frac{1}{\sqrt2}\right|\right]\right]$

$V_{avg}=\frac{2}{\pi}\left[\frac{10\pi}{4} \right ]=\frac{20}{4}=5$

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Example Question #169 : Calculus

Given the position function, calculate the velocity at t=2 .

p(t)=t^3-5t^2+10t-20

Possible Answers:

None of these

Correct answer:

Explanation:

Velocity is the derivative of position. The derivative of x^n is $nx^{n-1}$ .

That means that we can get the velocity fucntion with this information.

v(t)=3t^2-10t+10

To get the velocity at t=2 we must plug in 2 for t in the velocity function.

v(2)=3(2)^2-10(2)+10

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #161 : How To Find Velocity

Example Question #161 : How To Find Velocity

Example Question #162 : Calculus

Example Question #161 : Spatial Calculus

Example Question #161 : Velocity

Example Question #165 : Calculus

Example Question #161 : How To Find Velocity

Example Question #163 : How To Find Velocity

Example Question #164 : How To Find Velocity

Example Question #169 : Calculus

All Calculus 1 Resources

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