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Calculus 1 : Calculus

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Example Questions

Example Question #21 : Solving For Time

The position of a particle is described by the function r(t)=5t^2+7t-57 . How long does it take the particle to reach a speed of 15mph ?

Possible Answers:

t=0.8

t=10

t=1.2

t=-57

Correct answer:

t=0.8

Explanation:

We need to find when v(t)=15 . The velocity equation is the first derivative of the position equation. Taking the first derivative of the position equation gives

$v(t)=r'(t)=\frac{d}{dt}(5t^2)+\frac{d}{dt}(7t)-\frac{d}{dt}(57)$

v(t)=10t+7

Substituting v(t)=15 gives

15=10t+7

15-7=10t

8=10t

$t=\frac{8}{10}=0.8$

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Example Question #66 : Rate

The velocity of a satellite launched into space is described by the function v(t)=1600-100t . How long does it take the satellite to reach its highest point?

Possible Answers:

t=1600

t=1

t=100

t=16

Correct answer:

t=16

Explanation:

The rocket has reached it's highest point when v(t)=0 . Substituting this into the equation gives

0=1600-100t

Solving for , gives

1600=100t

$t=\frac{1600}{100}=16$

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Example Question #2891 : Calculus

The position of a bike is described by the function r(t)=2t^2-7t+6 . How long does it take the bike to reach a speed of 12 mph ?

Possible Answers:

t=1

t=19

t=5

t=4.75

Correct answer:

t=4.75

Explanation:

We need to find when v(t)=12 . The velocity equation is the first derivative of the position equation. Taking the first derivative of the position equation gives

$v(t)=r'(t)=\frac{d}{dt}(2t^2)-\frac{d}{dt}(7t)+\frac{d}{dt}(6)$

v(t)=4t-7

Substituting v(t)=12 gives

12=4t-7

19=4t

$t=\frac{19}{4}=4.75$

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Example Question #29 : Solving For Time

The velocity of a missle shot into the air is described by the function v(t)=400-56t . How long does it take the missile to hit the ground if it is launched from the ground?

Possible Answers:

t=100

$t=\frac{100}{7}$

t=50

t=7

Correct answer:

$t=\frac{100}{7}$

Explanation:

We need to find when r(t)=0 . To find the position equation, we need to integrate the velocity equation. Integrating the velocity equation gives

v(t)=400-56t

$\int v(t)=r(t)=\int (400-56t)dt$

$r(t)=400t-56(1/2)t^{2}+C$

$r(t)=400t-28t^{2}+C$

To find the constant , we use the inital conditon r(0)=0

$0=400(0)-28(0)^{2}+C$

0=C

Substituting in the constant

$r(t)=-28t^{2}+400t$

To find when r(t)=0 , we will use the quadratic equation

$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

In this equation, x=t , a=-28 , b=400 , and c=0 . Substituting these variables into quadratic equation gives

$t=\frac{-400\pm \sqrt{400^{2}-4\cdot(-28)\cdot(0)}}{2\cdot(-28)}$

$t=\frac{-400\pm \sqrt{400^{2}}}{-56}$

$t=\frac{-400\pm400}{-56}$

$t=\frac{-400+400}{-56}$ or $t=\frac{-400-400}{-56}$

The first answer gives us t=0 , which is when the rocket is launched. The second answer is when the missile hits the ground,

$t=\frac{800}{56}=\frac{100}{7}$

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Example Question #22 : Solving For Time

A ball thrown across the field at a parabolic rate described by the curve h=100-4t^2 . At what time will the ball hit the ground?

Possible Answers:

$\pm 5$

$\pm25$

Correct answer:

Explanation:

When the ball hits the ground, the height of the ball is zero. Substitute zero for the height and solve for .

0=100-4t^2

4t^2 = 100

t^2=25

$t=\pm 5$

Since there is no such thing as negative time, the ball will hit the ground after five seconds.

The answer is: t=5

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Example Question #31 : How To Solve For Time

A particle is accelerating at a rate given by the expression a(t) = 6t^2+3t+8 , where t is the time in seconds after the particle passes a gate inside of the particle accelerator. Two seconds after passing the gate, a radar detects that the particle was moving at a velocity of 50 m/s. The particle will continue to accelerate until it hits a lead wall found approximately 7,000 meters after passing the gate.

Find an expression in terms of t that gives the amount of time in seconds it takes for the particle to reach the wall after passing the gate.

Possible Answers:

0.5t^4+0.5t^3+4t^2-7000=0

t^4-0.5t^3+4t^2-12t=7000

t^4+t^3+8t^2+24t-7000=0

0.5t^4+0.5t^3+4t^2+12t-7000=0

2t^3+1.5t^2+8t+12=0

Correct answer:

0.5t^4+0.5t^3+4t^2+12t-7000=0

Explanation:

The acceleration of the particle is given by the follwoing formula: a(t) = 6t^2+3t+8 . In order to find the velocity of the particle, this need to be indefinetely integrated.

$v(t)=\int a(t)dt=\int (6t^2+3t+8)dt$

When taking an integral, you can seperate the integral into the sum of the integral of each part.

$\int(6t^2+3t+8)dt= \int6t^2dt+\int3tdt+\int8dt+C$

$\rightarrow \int6t^2dt=\frac{6}{3}t^3$

$\rightarrow\int 3tdt=\frac{3}{2}t^2$

$\rightarrow\int 8dt=8t$

Therefore, the expression for the velocity of the particle is determined to be:

v(t)=2t^3+1.5t^2+8t+C .

In order to find the constant C, you must use the conditions provided to you. The radar detects the velocity of the particle to be equal to 50 m/s two seconds after the particle passes the gate. Use this information to find the constant C.

$2t^3+1.5t^2+8t+C\rightarrow 2(2)^3+1.5(2)^2+8(2)+C= 50$

$\rightarrow16+6+16+C=50$

$\rightarrow C=12$

Now that the expression for velocity has been determined, the position function can be found by integrating the velocity function. The position function is used to find how long it takes for the particle to travel 7000 meters.

v(t)=2t^3+1.5t^2+8t+12

$p(t)=\int v(t)dt=\int (2t^3+1.5t^2+8t+12)dt$

Once again, this integral can be seperated into a sum of the integral of each part:

$\int v(t)dt=\int2t^3dt+\int1.5t^2dt+\int 8tdt+\int12dt+C$

$\rightarrow\int2t^3dt=\frac{2}{4}t^4$

$\rightarrow\int\frac{3}{2}t^2dt=\frac{3}{2*3}t^3$

$\rightarrow\int8tdt=\frac{8}{2}t^2$

$\rightarrow\int12dt=12t$

Since the start of the 7000 meters begins at the gate, the initial position of the particle is zero. Therefore, the constant C is equal to zero. This gives a position function for the particle:

p(t)=0.5t^4+0.5t^3+4t^2+12t

By setting this equal to the total distance traveled, the time t that it takes to reach 7000 meters can be determined.

7000=0.5t^4+0.5t^3+4t^2+12t

0.5t^4+0.5t^3+4t^2+12t-7000=0

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Example Question #31 : Solving For Time

A particle's velocity is modeled (in feet per second) by the equation:

y=4t^3-22t+1

At t=0 , the particle is at rest.

At what time is the particle's acceleration equal to 578 ft/s/s?

Possible Answers:

$t=\sqrt{35}$

$t=\sqrt{70}$

$t=5\sqrt{2}$

$8\sqrt{3}$

Correct answer:

$t=5\sqrt{2}$

Explanation:

For this question, we must first determine what the accleration function is by taking the derivative of the velocity equation:

Recall the power rule to differentiate.

The power rule states,

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

y=4t^3-22t+1

The equation for the acceleration of the particle, then, is:

y'=12t^2-22

With this equation in hand, we just have to solve for t when y'=578 :

578=12t^2-22

12t^2=600

t^2=50

$t=\sqrt{50}=5\sqrt{2}$

We are only interested in the positive solution to this problem, since the particle is not in motion before t=0.

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Example Question #32 : Solving For Time

Find how long it takes for a projectile starting at t=0 to hit a critical point if its height is given by:

h(t)=3tan(t)-sin(t)

Possible Answers:

$t=\frac{3\pi}{2}$

Never

$t=\frac{\pi}{3}$

$t=\pi$

Correct answer:

Never

Explanation:

To solve this, we have to take the first derivative and set it equal to zero:

Remember that derivative of tan(x)=sec^2(x) and derivative of sin(x)=cos(x)

h(t)=3tan(t)-sin(t)

$\frac{\mathrm{d} h}{\mathrm{d} t}= 3sec^2(t)-cos(t)=0$

Moving cos(t) to the other side,

$\frac{3}{cos^2(t)}=cos(t)$

We now get that:

cos^3(t)=3

From here, we can see that this equation will never equal zero since

$cos(t)=\sqrt[3]{3}$ will never have a solution since $\sqrt[3]{3}>1$ .

The correct answer is Never

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Example Question #34 : Solving For Time

There are equations in physics that can tell you about the position and velocity of an object as a function of time. If we were to graph the equation of the relationship between the object's height and time it would be a quadratic while graphing the velocity vs time would be linear (since velocity is the derivative of position in respect to time. The equation of the height of the projectile with respect to time is the following $h(t)=\frac{1}{2}at^2+v_0t+h_0$ where h_0 is the initial height, is the acceleration, and v_0 is the initial velocity.

Say you throw a ball straight up in the air at an initial velocity of 20m/s. At approximately what time does the ball reach it's highest height? Use $-9.8\ \frac{\textup{m}}{\textup{s}^2}$ as the approximation for the acceleration due to gravity.

Possible Answers:

1.02s

-1.02s

Not enough information given.

2.04s

Correct answer:

2.04s

Explanation:

Given that the relationship between height and time is a parabola, we note that the greatest height will occur when the first derivative is zero.

Given the equation,

$h(t)=\frac{1}{2}at^2+v_0t+h_0$

The derivative is

$v(t)=h'(t)=2\cdot \frac{1}{2}at^{2-1}+v_0t^{1-1}$

v(t)=at+v_0

which we then set equal to zero and plug in our known values.

We get the following:

0=-9.8 t+20 .

Then solving for time, t, gives us 2.04s.

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Example Question #71 : Rate

Find the average rate of change for x^2+x+1 over the interval x=[0, 5] .

Possible Answers:

31/5

Correct answer:

Explanation:

The rate of change of a function is the amount it changes over a given amount of time.

In mathematical terms, this can be written as

$r=\frac{\Delta y}{\Delta x}$

we plug in our values:

$\frac{5^2+5+1-0^2-0-1}{5-0} = \frac{30}{5}=6$

Note that this is only an average because quadratic functions change at different rates depending on where you are in the function's domain.

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #21 : Solving For Time

Example Question #66 : Rate

Example Question #2891 : Calculus

Example Question #29 : Solving For Time

Example Question #22 : Solving For Time

Example Question #31 : How To Solve For Time

Example Question #31 : Solving For Time

Example Question #32 : Solving For Time

Example Question #34 : Solving For Time

Example Question #71 : Rate

All Calculus 1 Resources

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