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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

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Example Questions

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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Example Question #2881 : Calculus

How much time does it take for a biker accelerating $5 \frac{m}{s^2}$ with initial velocity of $15 \frac{m}{s}$ , and initial position at to travel 2km ?

Possible Answers:

$\frac{-6+\sqrt{3200}}{1} s$

$\frac{-6+\sqrt{3236}}{2} s$

$\frac{-15+\sqrt{15286}}{5}s$

$\frac{-15+\sqrt{3236}}{5}s$

Correct answer:

$\frac{-6+\sqrt{3236}}{2} s$

Explanation:

First, recall that

$v(t)=\left[\int a(t)dt\right] +v_o$ ,

where v_o is the initial velocity, a(t) is the acceleration function, and v(t) is velocity.

By the power rule, we know that

$\int at^n dt= \frac{a}{n+1}t^{(n+1)} +C$ ,

where a, n, C are constants and is a variable.

In our case,

$v(t)=\left[\int 5dt\right]+15=5t+15$

Also recall that position is given as

$p(t)=\left[\int v(t)dt \right]+p_o$ ,

where p(t) is position at any given time and p_o is the initial position.

In our case, where p_o=0

$p(t)=\int [5t+15]dt= 2.5t^2+15t$ .

To travel 2000m , we set up the equation

2000=2.5t^2+15t .

This is equal to

t^2+6t-800=0

To solve this we use the quadratic formula, which states that for any quadratic equation:

Ax^2+Bx+C=0 , where A,B,C are constants, and is a variable

$x=\frac{-B+\sqrt{(B^2-4AC)}}{2A}$

Using the quadratic formula to solve t^2+6t-800=0 ,

$t=\frac{-6+\sqrt{(6^2-(-4*800))}}{2*1}= \frac{-6+\sqrt{3236}}{2}s$

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Example Question #51 : Rate

The position of car is described by the function r(t)=t^2-15t+30 . How long does it take the car to reach a speed of 65mph ?

Possible Answers:

t=40

t=20

t=400

t=50

Correct answer:

t=40

Explanation:

We need to find when v(t)=65 . The velocity equation is the first derivative of the position equation. Taking the first derivative of the position equation gives

$v(t)=r'(t)=\frac{d}{dt}(t^2)-\frac{d}{dt}(15t)+\frac{d}{dt}(30)$

v(t)=2t-15

Substituting v(t)=65 gives

65=2t-15

80=2t

t=40

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Example Question #11 : Solving For Time

The velocity of a ball thrown in the air is described by the function v(t)=32-32t . How long does it take the ball to reach its highest point?

Possible Answers:

t=32

t=100

t=1

t=0

Correct answer:

t=1

Explanation:

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

0=32-32t

Solving for , gives

32=32t

t=1

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Example Question #2884 : Calculus

The velocity of a bullet shot into the air is described by the function v(t)=350-5t . How long does it take the bullet to hit the ground if it starts above the ground?

Possible Answers:

t=350

$t=350+ \sqrt{122512}$

$t=350+ \sqrt{166}$

t=0

Correct answer:

$t=350+ \sqrt{122512}$

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

$\int v(t)=r(t)=\int (350-5t)dt$

$r(t)=350t-\frac{1}{2}t^{2}+C$

To find the constant , we use the inital conditon r(0)=6

$6=350(0)-\frac{1}{2}(0)^{2}+C$

6=C

Substituting in the constant

$r(t)=-\frac{1}{2}t^{2}+350t+6$

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=-1/2 , b=350 , and c=6 . Substituting these variables into quadratic equation gives

$t=\frac{-350\pm \sqrt{350^{2}-4*(-1/2)*(6)}}{2*(-1/2)}$

$t=\frac{-350\pm \sqrt{122500+12}}{-1}$

$t=350\mp \sqrt{122512}$

$t=350+ \sqrt{122512}$ or $t=350- \sqrt{122512}$

Since time cannot be negative, the answer is

$t=350+ \sqrt{122512}$

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Example Question #2885 : Calculus

The velocity of a ball is described by the function v(t)=98-4t . How long does it take the ball to hit the ground?

Possible Answers:

t=49

t=7

t=4

t=28

Correct answer:

t=49

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)=98-4t

$\int v(t)=r(t)=\int (98-4t)dt$

$r(t)=98t-4(1/2)t^{2}+C$

$r(t)=98t-2t^{2}+C$

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

$0=98(0)-2(0)^{2}+C$

0=C

Substituting in the constant

$r(t)=-2t^{2}+98t$

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=-2 , b=98 , and c=0 . Substituting these variables into quadratic equation gives

$t=\frac{-98\pm \sqrt{98^{2}-4*(-2)*(0)}}{2*(-2)}$

$t=\frac{-98\pm \sqrt{98^{2}}}{-4}$

$t=\frac{-98\pm 98}{-4}$

$t=\frac{-98+ 98}{-4}=0$ or $t=\frac{-98- 98}{-4}=49$

The first answer gives us t=0 , which is when the ball is thrown. The second answer is when the ball hits the ground,

t=49

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Example Question #21 : Solving For Time

The position of an car is described by the function r(t)=7t^2+5t-8 . How long does it take the car to travel 30ft ?

Possible Answers:

t=14

$t=\sqrt{1089}$

$t=\frac{ \sqrt{1089}}{14}$

$t=\frac{-5+ \sqrt{1089}}{14}$

Correct answer:

$t=\frac{-5+ \sqrt{1089}}{14}$

Explanation:

We need to find when r(t)=30ft .

30=7t^2+5t-8

0=7t^2+5t-38

To solve the above equation, use the quadratic equation

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

In this equation, x=t , a=7 , b=5 , and c=-38 . Substituting these variables into quadratic equation gives

$t=\frac{-5\pm \sqrt{5^{2}-4*7*(-38)}}{2*7}$

$t=\frac{-5\pm \sqrt{25+1064}}{14}$

$t=\frac{-5\pm \sqrt{1089}}{14}$

$t=\frac{-5+ \sqrt{1089}}{14}$ or $t=\frac{-5- \sqrt{1089}}{14}$

Since time cannot be negative, the answer is

$t=\frac{-5+ \sqrt{1089}}{14}$

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Example Question #22 : Solving For Time

Given the parametric equation for an object

x=3t

y=-t^3+3t^2

Determine at what the object reaches its maximum point. Assume t > 0 .

Possible Answers:

x=6

x=3

x=2

x=0

Correct answer:

x=6

Explanation:

The maximum point is reached when the derivative of with respect to time is .

y(t)=-t^3+3t^2

Using the power rule we know that:

$\frac{\mathrm{d} }{\mathrm{d} t}(at^n)=nat^{n-1}$ , where and are constants.

Therefore,

y'(t)=-3t^2+6t

Solving this for when y'(t)=0

-3t^2+6t=0

-3t(t-2)=0

The solutions are:

t=0 and t=2

Since the function is only defined for when t>0 , it hits its maximum at t=2 .

Since the problem asks for the value, we plug in t=2 for that parametric equation.

x=3*2=6

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Example Question #2886 : Calculus

The position of an airplane is described by the function r(t)=3t^2-12t+49 . How long does it take the airplane to travel 100m ?

Possible Answers:

t=122

$t=2+\frac{\sqrt{2}}{6}$

$t=2+\frac{ \sqrt{756}}{6}$

t=28849

Correct answer:

$t=2+\frac{ \sqrt{756}}{6}$

Explanation:

We need to find when r(t)=100m .

100=3t^2-12t+49

0=3t^2-12t+49-100

0=3t^2-12t-51

To solve the above equation, use the quadratic equation

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

In this equation, x=t , a=3 , b=-12 , and c=-51 . Substituting these variables into quadratic equation gives

$t=\frac{12\pm \sqrt{12^{2}-4\cdot3\cdot(-51)}}{2\cdot3}$

$t=\frac{12\pm \sqrt{144+612}}{6}$

$t=\frac{12\pm \sqrt{756}}{6}$

$t=2+\frac{ \sqrt{756}}{6}$ or $t=2-\frac{ \sqrt{756}}{6}$

Since time cannot be negative, the answer is

$t=2+\frac{ \sqrt{756}}{6}$

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Example Question #2887 : Calculus

The position of a person running is described by the function $r(t)=\frac{1}{2}t^2+t+6$ . How long does it take the person to run miles?

Possible Answers:

t=5

t=2

t=-4

t=16

Correct answer:

t=2

Explanation:

$r(t)=\frac{1}{2}t^2+t+6$

We need to find when r(t)=10 miles.

$10=\frac{1}{2}t^2+t+6$

$0=\frac{1}{2}t^2+t+6-10$

$0=\frac{1}{2}t^2+t-4$

To solve the above equation, use the quadratic equation

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

In this equation, x=t , a=1/2 , b=1 , and c=-4 . Substituting these variables into quadratic equation gives

$t=\frac{-1\pm \sqrt{1^{2}-4\cdot(1/2)\cdot(-4)}}{2\cdot(1/2)}$

$t=\frac{-1\pm \sqrt{1+8}}{1}$

$t=\frac{-1\pm \sqrt{9}}{1}$

t=-1+ 3=2 or t=-1- 3=-4

Time cannot be negative, so the answer is

t=2

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #53 : Rate

Example Question #2881 : Calculus

Example Question #51 : Rate

Example Question #11 : Solving For Time

Example Question #2884 : Calculus

Example Question #2885 : Calculus

Example Question #21 : Solving For Time

Example Question #22 : Solving For Time

Example Question #2886 : Calculus

Example Question #2887 : Calculus

All Calculus 1 Resources

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