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Calculus 1 : Calculus

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Example Questions

Example Question #4 : How To Solve For Time

Suppose the acceleration function of a biker going uphill at the start of a race is a(t)= 2t-8 , where is in seconds. When will it take the biker to reach constant velocity?

Possible Answers:

$2\sqrt2$

Correct answer:

Explanation:

Constant velocity means there is neither an increase or decrease in acceleration.

a=0

Substitute acceleration and solve for time.

a(t)= 2t-8

0=2t-8

t=4

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Example Question #5 : How To Solve For Time

For this problem, the acceleration of gravity is simplified to $10 \frac{m}{s^2}$ .

I fire a cannon straight into the air. Assuming that the cannonball leaves the cannon at a velocity of $100 \frac{m}{s}$ , weighs kgs, and is fired from the ground (i.e. m), how long will it take before the cannonball reaches the ground again?

Possible Answers:

$20\ seconds$

$10\ seconds$

$-10\ seconds$

$100 \frac{m}{s}$

$5\ seconds$

Correct answer:

$20\ seconds$

Explanation:

Initial velocity is given as 100 m/s and the acceleration due to gravity is $10\frac{m}{s^2}$ towards the Earth , or $-10 \frac{m}{s^2}$ .

We want to find how long it will take for the velocity of the cannonball to reach , so we set $v = 100 + a\cdot t = 0$ , where t is time and $a = -10 \frac{m}{s^2}$ .

So, 100 -10t = 0 , 100 = 10t , t = 10 .

Therefore, it takes ten seconds for the ball to reach a velocity of 0, and given that acceleration is uniform (i.e. a constant), we know that it will take the same amount of time to come down as it took to go up, or ten seconds. Therefore, the total time the cannonball spends in the air is,

10+10 = 20 seconds.

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Example Question #3 : How To Solve For Time

Given the instantaneous velocity and the position function, find the time at which the moving object reaches that instantaneous velocity.

v=125, p(t)=12t^2+29t-34

Possible Answers:

$t=\frac{1}{4}$

t=2.63

t=-5.04

t=4

Correct answer:

t=4

Explanation:

We begin by finding the derivative of the position function using the power rule:

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

p'(t)=v(t)=24t+29

We then set the given instantaneous velocity equal to the velocity function and solve for t:

$24t+29=125\Rightarrow t=4$

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Example Question #2 : How To Solve For Time

Assume the acceleration due to gravity is -9.8 m/s^2

If you throw a ball straight up with an initial velocity of 9.8 m/s how many seconds will it take before the ball returns to your hand?

Possible Answers:

0.64

0.5

1.5

Correct answer:

Explanation:

Write out the equation for the height of the ball

$h=-\frac{9.8}{2}t^{2}+9.8t$ .

You can arrive at this by starting with the information that the acceleration on the ball is the constant acceleration due to gravity and integrating twice. Making sure to solve for your constants along the way.

$v(t)=\int a(t)dt = \int -9.8 dt = -9.8t+v_{0}$

$x(t)=\int v(t)dt =\int -9.8t+v_{0}dt=-\frac{9.8}{2}t^{2}+v_{0}t+x_{0}$

Your initial velocity of 9.8m/s and your initial position of will help write out this equation.

$x(t)=-\frac{9.8}{2}t^2+9.8t+0$

Then solve for the two values of for which the ball is at height . Those are seconds and seconds.

$\\x(t)=-\frac{9.8}{2}t^2+9.8t+0=0\\ \\ -4.9t^{2}+9.8t=0\\ \\ t(-4.9t+9.8)=0\\ \\t=0,t=2$

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Example Question #3 : How To Solve For Time

A ball is thrown upwards at a speed of 10m/s from a 100 m building. Assume gravity is 9.8m/s .

Which of the following is closest to the time after the initial throw before the ball hits the ground?

Possible Answers:

$1+\sqrt{21}$

$1+\sqrt{12}$

Correct answer:

$1+\sqrt{21}$

Explanation:

If we approximate gravity as -10m/s^2 we can simplify 0=-5t^2+10t+100 into 0=-t^2+2t+20 and use the quadratic formula to find the time at which the position of the ball is zero (the ball hits the ground).

$t=\frac{-2\pm \sqrt{(2)^2-4(-1)(20)}}{2(-1)}=1\pm\frac{\sqrt{4+80}}{-2}=1\mp\sqrt{21}$

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Example Question #11 : How To Solve For Time

At what time will a particle whose position can be described by have minimum acceleration?

Possible Answers:

$\small -0.27$

$\small 0.40$

$\small 0.27$

$\small -0.52$

$\small 0.12$

Correct answer:

$\small 0.27$

Explanation:

We need to take the derivative of acceleration and set it equal to zero because we want to minimize acceleration. In total, this will be three derivatives

$\small x(t)\rightarrow v(t) \rightarrow a(t)\rightarrow \frac{\mathrm{d} }{\mathrm{d} t}a(t)$ .

Using the power rule on each term which states to multiply the coefficient by the exponent then decrease the exponent by one we get the following derivatives.

v(t)=20t^4-9t^2+2t

a(t)=80t^3-18t+2

a'(t)=240t^2-18

This will give us

240t^2-18=0 , which gives us $\small t=.27$ and $\small t=-.27$ . Now let's use the second derivative test to see where our minimum is. $\small 480t$ is our second derivative, which is positive at $\small t=.27$ and negative at $\small t=-.27$ , so $\small t=.27$ is our minimum.

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Example Question #51 : Rate

A perfectly spherical hot air balloon is being filled up. If the balloon is empty at the start and has a radius of 50 meters when fully inflated, how fast is the volume of the balloon increasing when its radius is 10 meters and increasing at a rate of $1\frac{m}{min}$ ?

Possible Answers:

None of the above.

$100 \pi \frac{m}{min}$

$400 \pi \frac{m}{min}$

$300 \pi \frac{m}{min}$

$200 \pi \frac{m}{min}$

Correct answer:

$400 \pi \frac{m}{min}$

Explanation:

In order to solve this problem, we must first know that the volume of a sphere is equivalent to $\frac{4}{3}\pi r^3$ .

In order to find the rate of which the volume of this spherical hot air balloon is increasing at, we must take the derivative of the volume equation with respect to time in order to find the change in volume with respect to time.

Using the power rule

$\frac{d}{dt}x^n=nx^{n-1}$ ,

we find that the dervative is

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$ .

In the problem we are given the radius and rate of change of the radius, therefore by plugging those into the equation and solving for $\frac{dV}{dt}$ , we can find the rate at which the volume of the balloon is changing at.

Plugging r=10 and $\frac{dr}{dt}=1$ , we find that the volume of the baloon is increasing at a rate of $400 \pi \frac{m}{min}$ .

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Example Question #13 : How To Solve For Time

Car A starts driving north from point O with an acceleration of 10 mph . After 2 hours, Car B start driving north from point O with an acceleration of 30 mph . How long will it take for Car B to catch up with Car A?

Possible Answers:

$t=\sqrt{2} hours$

t=1.5 hours

None of the above.

$t=1+\sqrt{3} hours$

$t=\sqrt{3} hours$

Correct answer:

$t=1+\sqrt{3} hours$

Explanation:

We know that car A's acceleration formula is $a_{A}(t)=10$ and we know that car B's acceleration formula is $a_{B}(t)=30$ . To solve this equation we must realize that the integral of acceleration is velocity and the integral of velocity is position. Therefore by taking the double integral of both acceleration functions, we can determine the point at which car B will catch up to car A.

Using the general integral formula,

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

we find that the velocity functions for both cars are $v_{A}(t)=10t$ and $v_{B}(t)=30t$ . Because the initial velocity of both cars is 0, C=0 .

Taking the integral of the velocity function using the generla integral formula once again, we find that the position functions of both cars is $S_{A}(t)=5t^2$ and $S_{B}(t)=15t^2$ . Since the initial position of both cars is equivalent, we can arbitrarily say that they start from a initial position 0, therefore making C=0 . We know that car A had a head start of 2 hours on car B. Now all we have to do is set both equations equal to each other and solve for .

5(t+2)^2+20=15t^2

5t^2+20t+20=15t^2

t^2-2t-2=0

$t=1+\sqrt{3} hours$

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Example Question #51 : Rate

Given the position function, at what time is the velocity going to be equal to zero?

p(t)=t^3-15t^2+27t-7

Possible Answers:

None of these

t=1,9

t=1

t=2,7

t=9

Correct answer:

t=1,9

Explanation:

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

Using this information we can find te velocity function.

v(t)=3t^2-30t+27

To find where the velocity is 0, we msut set the velocity function to 0 and factor to solve.

3(t^2-10t+9)=0

t^2-10t+9=0

(t-9)(t-1)=0

t=1,9

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Example Question #53 : Rate

The position of a particle as a function of time is s(t)=4t^3-t^4 .

At what time t>0 is the particle at rest?

Possible Answers:

Correct answer:

Explanation:

The particle is at rest when its velocity, i.e. the derivative of its position, is equal to 0.

Thus, we have to solve the equation

v(t)=s'(t)=0 .

Using the Power Rule,

s'(t)=12t^2-4t^3=t^2(12-4t) .

Thus, either t^2=0 or 12-4t=0 , leading to the solutions t=0 and t=3 .

Note: The Power Rule says that for a function

p(x)=kx^n , $p'(x)=knx^{n-1}$ .

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #4 : How To Solve For Time

Example Question #5 : How To Solve For Time

Example Question #3 : How To Solve For Time

Example Question #2 : How To Solve For Time

Example Question #3 : How To Solve For Time

Example Question #11 : How To Solve For Time

Example Question #51 : Rate

Example Question #13 : How To Solve For Time

Example Question #51 : Rate

Example Question #53 : Rate

All Calculus 1 Resources

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