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Calculus 1 : Rate

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Example Question #20 : How To Solve For Time

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=28

t=4

t=7

t=49

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 #21 : How To Solve For Time

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:

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

t=122

t=28849

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

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 #24 : Solving For Time

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=2

t=16

t=-4

t=5

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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Example Question #25 : Solving For Time

The velocity of a rocket shot into the air is described by the function v(t)=100-9.8t . How long does it take the rocket to reach its highest point?

Possible Answers:

t=10.2

t=100

t=9.8

t=1.5

Correct answer:

t=10.2

Explanation:

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

0=100-9.8t

Solving for , gives

100=9.8t

$t=\frac{100}{9.8}=10.2$

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Example Question #26 : 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=1.2

t=-57

t=10

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 #27 : Solving For Time

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=1

t=1600

t=16

t=100

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 #28 : Solving For Time

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=4.75

t=19

t=1

t=5

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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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #20 : How To Solve For Time

Example Question #21 : Solving For Time

Example Question #22 : Solving For Time

Example Question #21 : How To Solve For Time

Example Question #24 : Solving For Time

Example Question #25 : Solving For Time

Example Question #26 : Solving For Time

Example Question #27 : Solving For Time

Example Question #28 : Solving For Time

Example Question #29 : Solving For Time

All Calculus 1 Resources

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