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Calculus 1 : How to solve for time

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Example Questions

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 #11 : Solving For Time

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:

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

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

None of the above.

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

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

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 #11 : How To Solve For Time

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:

t=1,9

t=1

t=9

t=2,7

None of these

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 #1851 : Functions

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 #11 : How To Solve For Time

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=20

t=50

t=400

t=40

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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Calculus 1 : How to solve for time

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #11 : How To Solve For Time

Example Question #11 : Solving For Time

Example Question #13 : How To Solve For Time

Example Question #11 : How To Solve For Time

Example Question #1851 : Functions

Example Question #2881 : Calculus

Example Question #11 : How To Solve For Time

Example Question #11 : Solving For Time

Example Question #2884 : Calculus

Example Question #2885 : Calculus

All Calculus 1 Resources

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