Calculus using Calculus using Distance, Velocity Distance, Velocity and Accelerationand Acceleration

By Tony FarahBy Tony Farah

Distance (Displacement)Distance (Displacement)

Distance can be defined as the Distance can be defined as the difference between point a and difference between point a and point b as a function of timepoint b as a function of time

An example would be that a An example would be that a baseball is hit and falls just shy baseball is hit and falls just shy of homerunof homerun

The displacement, or its The displacement, or its distance above the ground, is distance above the ground, is from the time the ball hits the from the time the ball hits the bat until the ball falls into bat until the ball falls into center field center field

It can be defined in the It can be defined in the following function where y is following function where y is displacement in feet above the displacement in feet above the ground and t is the number of ground and t is the number of seconds since the ball was hitseconds since the ball was hit

Distance is measured in feetDistance is measured in feet

y (t) = -14ty (t) = -14t22 + 23t + 2 + 23t + 2

VelocityVelocity

y (t) = -14ty (t) = -14t22 + 23t + 2 + 23t + 2

y ’ (t) = -28t + 23y ’ (t) = -28t + 23

Velocity can be defined Velocity can be defined as how fast an object is as how fast an object is going and in what going and in what directiondirection

Velocity can also be Velocity can also be defined as an defined as an instantaneous rate of instantaneous rate of change or a change or a derivativederivative

In this case the velocity In this case the velocity is how fast the ball is hit is how fast the ball is hit and in what direction it and in what direction it travelstravels

Velocity = dy/dt or y ’ (t) Velocity = dy/dt or y ’ (t) (y prime of t)(y prime of t)

Velocity is in ft/secVelocity is in ft/sec

AccelerationAcceleration

Acceleration occurs Acceleration occurs when velocity changeswhen velocity changes

It is the derivative of It is the derivative of velocity and the 2velocity and the 2ndnd derivative of derivative of displacementdisplacement

Negative acceleration Negative acceleration means velocity is means velocity is decreasingdecreasing

It can be defined as It can be defined as dv/dt or y ’’ (t) (y double dv/dt or y ’’ (t) (y double prime of t)prime of t)

Acceleration is in Acceleration is in ft/sec/secft/sec/sec

y (t) = -14ty (t) = -14t22 + 23t + 2 + 23t + 2

y ’ (t) = -28t + 23y ’ (t) = -28t + 23

y ” (t) = -28y ” (t) = -28

Key InformationKey Information

To find velocity at a particular time, substitute for t in y To find velocity at a particular time, substitute for t in y (t)(t)

Negative velocity means distance is decreasingNegative velocity means distance is decreasing If velocity and acceleration have the same sign, speed If velocity and acceleration have the same sign, speed

increases and vice versaincreases and vice versa Negative acceleration means velocity is decreasingNegative acceleration means velocity is decreasing Distance is measured in ft, velocity in ft/sec, and Distance is measured in ft, velocity in ft/sec, and

acceleration in ft/sec/secacceleration in ft/sec/sec A change in direction can be indicated when a graph A change in direction can be indicated when a graph

goes from negative to positive values or from positive goes from negative to positive values or from positive to negative valuesto negative values

Example Number 1Example Number 1

Jane and some friends are playing kickballJane and some friends are playing kickballafter school. Jane kicks the ball so hard itafter school. Jane kicks the ball so hard itflies up and over the fence. As it rises andflies up and over the fence. As it rises and

falls its distance above the ground is afalls its distance above the ground is afunction of time since the kick measured in feet.function of time since the kick measured in feet.

The function can be defined as f (t) = -30tThe function can be defined as f (t) = -30t33 + 12t + 12t22 + 8t +2 + 8t +2

a)a) Find velocity and determine what velocity is at t = 1Find velocity and determine what velocity is at t = 1b)b) Is the distance increasing or decreasing at t = 1 and Is the distance increasing or decreasing at t = 1 and

explainexplainc)c) Determine acceleration of the function and determine Determine acceleration of the function and determine

acceleration at t=1acceleration at t=1

Part APart A

Velocity is the derivative of acceleration soVelocity is the derivative of acceleration sotake the function’s derivativetake the function’s derivative

f (t) = -30tf (t) = -30t33 + 12t + 12t22 + 8t +2 + 8t +2f ’ (t) = -90tf ’ (t) = -90t22 + 24t + 8 + 24t + 8

Now since we know velocity we can plugNow since we know velocity we can pluginto the equation for t.into the equation for t.

f’ (1)= -90(1)f’ (1)= -90(1)22 + 24(1)+ 8 + 24(1)+ 8f’ (1)= -58 feet/secf’ (1)= -58 feet/sec

Part BPart BPart BPart B

Now since we know velocity = -58 ft/sec we canNow since we know velocity = -58 ft/sec we can

determine what this means in terms ofdetermine what this means in terms of

distance. The velocity here is decreasingdistance. The velocity here is decreasing

because the its sign is negative at t = 1.because the its sign is negative at t = 1.

Whenever velocity is equals a negativeWhenever velocity is equals a negative

number its distance is always decreasing. From number its distance is always decreasing. From

this we can also concur that positive velocity this we can also concur that positive velocity

means that distance is increasing.means that distance is increasing.

Part CPart C

We can find acceleration by just taking theWe can find acceleration by just taking thederivative of velocity.derivative of velocity.

f ’ (t) = -90tf ’ (t) = -90t22 + 24t + 8 + 24t + 8f ” (t) = -180t + 24f ” (t) = -180t + 24

Now we need to find acceleration at t = 1Now we need to find acceleration at t = 1

f ” (1) = -180(1) + 24f ” (1) = -180(1) + 24

f ” (1) = -156 ft/sec/secf ” (1) = -156 ft/sec/sec

Example Number 2Example Number 2

Find the averageFind the average

velocity at t = 3 byvelocity at t = 3 by

using the average rateusing the average rate

of changeof change

Hint: Hint: y2 – y1y2 – y1

x2 – x 1x2 – x 1

00 1010

22 44

44 1717

66 2121

t d (t)

How To Solve ItHow To Solve It

To use this the formula we can pick anyTo use this the formula we can pick anytwo y values and any two x values and findtwo y values and any two x values and findits slope or average rate of change. The bestits slope or average rate of change. The bestestimate would be to use the values closest to estimate would be to use the values closest to t = 3t = 3

17 – 417 – 4 = = 13134 – 2 24 – 2 2

This answer is the average velocity at t = 3This answer is the average velocity at t = 3

Multiple Choice # 1 Multiple Choice # 1 (No Calculator)(No Calculator)

A bug begins to crawl up a vertical wire at time t = 0. The velocity v of theA bug begins to crawl up a vertical wire at time t = 0. The velocity v of thebug at time t, 0 < t < 8, is given by the function whose graph is shown above.bug at time t, 0 < t < 8, is given by the function whose graph is shown above.

At what value of t does the bug change direction?At what value of t does the bug change direction?

a.) 2a.) 2 b.) 4b.) 4 c.) 6c.) 6 d.) 7 d.) 7 e.) 8e.) 8

Answer key to Multiple Answer key to Multiple Choice # 1Choice # 1

Answer = c.) 6Answer = c.) 6

The reason for this is that the graph goesThe reason for this is that the graph goes

or changes from positive to negative valuesor changes from positive to negative values

at t = 6 indicating a change in direction.at t = 6 indicating a change in direction.

Multiple Choice # 2Multiple Choice # 2(with calculator)(with calculator)

    s (t) = s (t) = tt22 - 20 - 20

Find the average velocity from t = 3 to t = 5Find the average velocity from t = 3 to t = 5

a.) 2a.) 2 b.)4b.)4 c.)6c.)6 d.)8d.)8

Answer Key to Multiple Answer Key to Multiple Choice #2Choice #2

For this problem we must use the averageFor this problem we must use the average

velocity formulavelocity formula

s(5) – s(3)s(5) – s(3) = = 5 – (-11)5 – (-11) = = 1616 = 8 = 8

5 – 3 2 25 – 3 2 2

Choice d.) 8Choice d.) 8

Multiple Choice # 3Multiple Choice # 3(no calculator)(no calculator)

An objects distance from its starting pointAn objects distance from its starting point

at time (t) is given by the equation at time (t) is given by the equation

d (t) = td (t) = t3 3 - 6t- 6t2 2 - 4. - 4. What is the speed ofWhat is the speed of

the object when its acceleration is 0?the object when its acceleration is 0?

a.) 2a.) 2 b.) -24 c.) 22 b.) -24 c.) 22 d.) 44 d.) 44 e.) -12 e.) -12

Answer Key to Multiple Answer Key to Multiple Choice # 3Choice # 3

Choice a.) 2Choice a.) 2

You must take the derivative of distanceYou must take the derivative of distancethen the derivative of velocitythen the derivative of velocity

d’ (t) = 3d’ (t) = 3tt22 - 12t - 12t d” (t) = 6t – 12d” (t) = 6t – 12

Now you must solve for t by setting d” orNow you must solve for t by setting d” oracceleration = 0acceleration = 0

6t – 12 = 06t – 12 = 0 = = 6t = 126t = 12 t = 2 t = 2 +12 +12+12 +12 6 6 6 6

Multiple Choice # 4Multiple Choice # 4

A particle moves along the x-axis so that itsA particle moves along the x-axis so that its

position at time t is given by position at time t is given by

d (t) = td (t) = t22 – 6t + 5. For what value of t is the – 6t + 5. For what value of t is the

velocity of the particle zero?velocity of the particle zero?

a.) 1a.) 1 b.) 2b.) 2 c.) 3c.) 3 d.) 4d.) 4e.) 5e.) 5

Answer Key to Multiple Answer Key to Multiple choice # 4choice # 4

Choice a.) 3Choice a.) 3

First we find the derivative of the distanceFirst we find the derivative of the distancethan we set = 0 just like multiple choicethan we set = 0 just like multiple choiceproblem # 3problem # 3

V(t) = 2t – 6V(t) = 2t – 6 2t - 6 = 02t - 6 = 0 2t = 62t = 6 t = 3 t = 3 +6+6 +6 +6 2 22 2

Free Response # 1 (you Free Response # 1 (you may use your calculator)may use your calculator)

Professor Frink never anticipated Truckasaurus getting loose, but then again, he never Professor Frink never anticipated Truckasaurus getting loose, but then again, he never foresaw the obvious dangers in reanimating the corpse of his long-dead father. Frinkie's foresaw the obvious dangers in reanimating the corpse of his long-dead father. Frinkie's not one for long-term planning. Anyway, Homer Simpson lies directly in the path of the not one for long-term planning. Anyway, Homer Simpson lies directly in the path of the flame-spewing juggernaut, with only the meager acceleration of the family station flame-spewing juggernaut, with only the meager acceleration of the family station wagon standing between him and utter destruction. wagon standing between him and utter destruction.

Assume Homer's velocity (in feet per second) is given by the equation , where Assume Homer's velocity (in feet per second) is given by the equation , where tt is is measured in seconds and . Answer the following questions based on the given measured in seconds and . Answer the following questions based on the given information, accurate to the thousandths place. information, accurate to the thousandths place.

(a) Is Homer traveling backwards at any time during the first 7 seconds of his escape (a) Is Homer traveling backwards at any time during the first 7 seconds of his escape attempt? If so, during what time interval(s) is the gear shift in reverse?attempt? If so, during what time interval(s) is the gear shift in reverse?

(b) What is Homer's average velocity from (b) What is Homer's average velocity from tt = 2 to = 2 to tt = 5? = 5?

(c) At what time(s) does Homer change from accelerating to decelerating, or vice versa?(c) At what time(s) does Homer change from accelerating to decelerating, or vice versa?

Answer to Free Answer to Free Response # 1Response # 1

(a) Homer is traveling backward whenever his velocity is negative. In other words, (a) Homer is traveling backward whenever his velocity is negative. In other words, you're looking for the intervals of you're looking for the intervals of vv((tt) when its graph falls below the ) when its graph falls below the xx-axis. If you -axis. If you graph graph vv on your calculator, you'll find 2 on your calculator, you'll find 2 xx-intercepts in the interval [0,7]. It's -intercepts in the interval [0,7]. It's advisable to use the calculator to find them: advisable to use the calculator to find them: xx = .40642065 and = .40642065 and xx = 4.1819433. = 4.1819433. Since the graph is below the Since the graph is below the xx-axis between those -axis between those xx-intercepts, Homer's in reverse -intercepts, Homer's in reverse on the interval (.406, 4.182). on the interval (.406, 4.182).

(b) Since you're given the velocity function, and are asked to find its average value, (b) Since you're given the velocity function, and are asked to find its average value, you have to use the average value formula, which states that the average value of you have to use the average value formula, which states that the average value of ff on the interval [on the interval [aa,,bb] is . For this problem, ] is . For this problem, aa = 2 and = 2 and bb = 5. Since you're allowed to = 5. Since you're allowed to use a calculator, it would be a waste of time to compute the antiderivative by hand. use a calculator, it would be a waste of time to compute the antiderivative by hand. The answer is -3.75. The answer is -3.75.

(c) Homer will change from acceleration to deceleration (or vice versa) whenever the(c) Homer will change from acceleration to deceleration (or vice versa) whenever thederivative of derivative of ff has an has an xx-intercept (since the derivative of the velocity function is the-intercept (since the derivative of the velocity function is theacceleration function). (Note that acceleration function). (Note that ff also must cross the axis there, as well, not also must cross the axis there, as well, not simplysimplybounce off of it, or there will be no change in the sign of the graph.) The derivativebounce off of it, or there will be no change in the sign of the graph.) The derivativeof of ff is . The acceleration function has only one is . The acceleration function has only one xx-intercept on the interval [0,7]: -intercept on the interval [0,7]: xx = =2.786 seconds, at which Homer goes from decelerating to accelerating. 2.786 seconds, at which Homer goes from decelerating to accelerating.

Free Response # 2Free Response # 2

A particle moves along the y-axis with the velocity givenA particle moves along the y-axis with the velocity givenby v (t) = t sin ( tby v (t) = t sin ( t22 ) for t is greater than or equal to 0. ) for t is greater than or equal to 0.

a.) In which direction (up or down) is the particle moving a.) In which direction (up or down) is the particle moving at time t = 1.5? Why?at time t = 1.5? Why?

b.) Find the acceleration of the particle at time t = 1.5. Is b.) Find the acceleration of the particle at time t = 1.5. Is the velocity of the particle increasing at t = 1.5? Why or the velocity of the particle increasing at t = 1.5? Why or why not?why not?

c.) Given that y (t) is the position of the particle at time t c.) Given that y (t) is the position of the particle at time t and that y (0) = 3, find y (2).and that y (0) = 3, find y (2).

d.) Find the total distance traveled by the particle from t = d.) Find the total distance traveled by the particle from t = 0 to t = 2.0 to t = 2.

Answer Key to Free Answer Key to Free Response # 2Response # 2

a.) v (1.5) = 1.5sin(1.5a.) v (1.5) = 1.5sin(1.522) = 1.167) = 1.167

Up, because v (1.5) is greater than 0Up, because v (1.5) is greater than 0

b.) a (t) = v’ (t) = sin tb.) a (t) = v’ (t) = sin t22 + 2 t + 2 t22 cos t cos t22

a (1.5) = v’ (1.5) = -2.048 or -2.049a (1.5) = v’ (1.5) = -2.048 or -2.049

No; v is decreasing at 1.5 because v’ (1.5)No; v is decreasing at 1.5 because v’ (1.5)

is less than 0is less than 0

Continued Answer KeyContinued Answer Key

c.) y(t) = the integral of v(t) dt c.) y(t) = the integral of v(t) dt = the integral of tsin t= the integral of tsin t22 dt = dt = - cos t- cos t22 + C + C

22y (0) = 3 = -y (0) = 3 = -11+ C = + C = 77 2 22 2

y (t) = - y (t) = - 11cos tcos t2 2 + + 77 2 22 2

y (2) = - y (2) = - 11cos4 + cos4 + 7 7 = 3.826 or 3.827= 3.826 or 3.827 2 22 2

d.) distance = the integral from 0 to 2 of thed.) distance = the integral from 0 to 2 of theabsolute value of v(t) = 1.173absolute value of v(t) = 1.173

Works CitedWorks Cited

Mr. Watson Class Notes and WorksheetsMr. Watson Class Notes and Worksheets

““Average velocity.” Average velocity.” Chapter 20-4 Velocity and Chapter 20-4 Velocity and AccelerationAcceleration. 7 June 2007 . 7 June 2007 <http://home.alltel.net/ okrebs/ page205.html>.<http://home.alltel.net/ okrebs/ page205.html>.

Foerster, Paul A. Foerster, Paul A. CalculusCalculus. EmeryVille: Steve . EmeryVille: Steve Rasmussen, 1998.Rasmussen, 1998.

““Problem of the Year 2003-2004.” Problem of the Year 2003-2004.” Calculus-HelpCalculus-Help. 7 June . 7 June 2007 <http://www.calculus-help.com/>.2007 <http://www.calculus-help.com/>.