Calculus using Distance, Velocity and Acceleration

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Calculus using Distance, Velocity and Acceleration

• By Tony Farah

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Distance (Displacement)

• Distance can be defined as the difference between
  point a and point b as a function of time
• An example would be that a baseball is hit and
  falls just shy of homerun
• The displacement, or its distance above the
  ground, is from the time the ball hits the bat
  until the ball falls into center field
• It can be defined in the following function where
  y is displacement in feet above the ground and t
  is the number of seconds since the ball was hit
• Distance is measured in feet

• y (t) -14t2 23t 2

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Velocity

• y (t) -14t2 23t 2
• y (t) -28t 23

• Velocity can be defined as how fast an object is
  going and in what direction
• Velocity can also be defined as an instantaneous
  rate of change or a derivative
• In this case the velocity is how fast the ball is
  hit and in what direction it travels
• Velocity dy/dt or y (t) (y prime of t)
• Velocity is in ft/sec

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Acceleration

• Acceleration occurs when velocity changes
• It is the derivative of velocity and the 2nd
  derivative of displacement
• Negative acceleration means velocity is
  decreasing
• It can be defined as dv/dt or y (t) (y double
  prime of t)
• Acceleration is in ft/sec/sec

• y (t) -14t2 23t 2
• y (t) -28t 23
• y (t) -28

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Key Information

• To find velocity at a particular time, substitute
  for t in y (t)
• Negative velocity means distance is decreasing
• If velocity and acceleration have the same sign,
  speed increases and vice versa
• Negative acceleration means velocity is
  decreasing
• Distance is measured in ft, velocity in ft/sec,
  and acceleration in ft/sec/sec
• A change in direction can be indicated when a
  graph goes from negative to positive values or
  from positive to negative values

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Example Number 1

• Jane and some friends are playing kickball
• after school. Jane kicks the ball so hard it
• flies up and over the fence. As it rises and
• falls its distance above the ground is a
• function of time since the kick measured in feet.
• The function can be defined as f (t) -30t3
  12t2 8t 2
• Find velocity and determine what velocity is at t
  1
• Is the distance increasing or decreasing at t 1
  and explain
• Determine acceleration of the function and
  determine acceleration at t1

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Part A

• Velocity is the derivative of acceleration so
• take the functions derivative
• f (t) -30t3 12t2 8t 2
• f (t) -90t2 24t 8
• Now since we know velocity we can plug
• into the equation for t.
• f (1) -90(1)2 24(1) 8
• f (1) -58 feet/sec

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Part B

• Now since we know velocity -58 ft/sec we can
• determine what this means in terms of
• distance. The velocity here is decreasing
• because the its sign is negative at t 1.
• Whenever velocity is equals a negative
• number its distance is always decreasing. From
• this we can also concur that positive velocity
• means that distance is increasing.

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Part C

• We can find acceleration by just taking the
• derivative of velocity.
• f (t) -90t2 24t 8
• f (t) -180t 24
• Now we need to find acceleration at t 1
• f (1) -180(1) 24
• f (1) -156 ft/sec/sec

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Example Number 2
t d (t)

• Find the average
• velocity at t 3 by
• using the average rate
• of change
• Hint y2 y1
• x2 x 1

0 10
2 4
4 17
6 21
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How To Solve It

• To use this the formula we can pick any
• two y values and any two x values and find
• its slope or average rate of change. The best
• estimate would be to use the values closest to
• t 3
• 17 4 13
• 4 2 2
• This answer is the average velocity at t 3

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Multiple Choice 1 (No Calculator)

• A bug begins to crawl up a vertical wire at time
  t 0. The velocity v of the
• bug at time t, 0 lt t lt 8, is given by the
  function whose graph is shown above.
• At what value of t does the bug change direction?
• a.) 2 b.) 4 c.) 6 d.) 7 e.) 8

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Answer key to Multiple Choice 1

• Answer c.) 6
• The reason for this is that the graph goes
• or changes from positive to negative values
• at t 6 indicating a change in direction.

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Multiple Choice 2(with calculator)

•   s (t) t2 - 20
• Find the average velocity from t 3 to t 5
• a.) 2 b.)4 c.)6 d.)8

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Answer Key to Multiple Choice 2

• For this problem we must use the average
• velocity formula
• s(5) s(3) 5 (-11) 16 8
• 5 3 2 2
• Choice d.) 8

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Multiple Choice 3(no calculator)

• An objects distance from its starting point
• at time (t) is given by the equation
• d (t) t3 - 6t2 - 4. What is the speed of
• the object when its acceleration is 0?
• a.) 2 b.) -24 c.) 22 d.) 44
  e.) -12

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Answer Key to Multiple Choice 3

• Choice a.) 2
• You must take the derivative of distance
• then the derivative of velocity
• d (t) 3t2 - 12t
• d (t) 6t 12
• Now you must solve for t by setting d or
• acceleration 0
• 6t 12 0 6t 12 t 2
• 12 12 6 6

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Multiple Choice 4

• A particle moves along the x-axis so that its
• position at time t is given by
• d (t) t2 6t 5. For what value of t is the
• velocity of the particle zero?
• a.) 1 b.) 2 c.) 3 d.) 4 e.) 5

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Answer Key to Multiple choice 4

• Choice a.) 3
• First we find the derivative of the distance
• than we set 0 just like multiple choice
• problem 3
• V(t) 2t 6 2t - 6 0 2t 6 t 3
• 6 6 2 2

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Free Response 1 (you may use your calculator)

• 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 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 wagon standing between him and
  utter destruction.
• Assume Homer's velocity (in feet per second) is
  given by the equation , where t is measured in
  seconds and . Answer the following questions
  based on the given information, accurate to the
  thousandths place.
• (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?
• (b) What is Homer's average velocity from t 2
  to t 5?
• (c) At what time(s) does Homer change from
  accelerating to decelerating, or vice versa?

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Answer to Free Response 1

• (a) Homer is traveling backward whenever his
  velocity is negative. In other words, you're
  looking for the intervals of v(t) when its graph
  falls below the x-axis. If you graph v on your
  calculator, you'll find 2 x-intercepts in the
  interval 0,7. It's advisable to use the
  calculator to find them x .40642065 and x
  4.1819433. Since the graph is below the x-axis
  between those x-intercepts, Homer's in reverse on
  the interval (.406, 4.182).
• (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 f on the interval a,b is
  . For this problem, a 2 and b 5. Since you're
  allowed to use a calculator, it would be a waste
  of time to compute the antiderivative by hand.
  The answer is -3.75.
• (c) Homer will change from acceleration to
  deceleration (or vice versa) whenever the
• derivative of f has an x-intercept (since the
  derivative of the velocity function is the
• acceleration function). (Note that f also must
  cross the axis there, as well, not simply
• bounce off of it, or there will be no change in
  the sign of the graph.) The derivative
• of f is . The acceleration function has only
  one x-intercept on the interval 0,7 x
• 2.786 seconds, at which Homer goes from
  decelerating to accelerating.

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Free Response 2

• A particle moves along the y-axis with the
  velocity given
• by v (t) t sin ( t2 ) for t is greater than or
  equal to 0.
• a.) In which direction (up or down) is the
  particle moving at time t 1.5? Why?
• 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 why not?
• c.) Given that y (t) is the position of the
  particle at time t and that y (0) 3, find y
  (2).
• d.) Find the total distance traveled by the
  particle from t 0 to t 2.

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Answer Key to Free Response 2

• a.) v (1.5) 1.5sin(1.52) 1.167
• Up, because v (1.5) is greater than 0
• b.) a (t) v (t) sin t2 2 t2 cos t2
• a (1.5) v (1.5) -2.048 or -2.049
• No v is decreasing at 1.5 because v (1.5)
• is less than 0

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Continued Answer Key

• c.) y(t) the integral of v(t) dt
• the integral of tsin t2 dt - cos t2 C
• 2
• y (0) 3 -1 C 7
• 2 2
• y (t) - 1cos t2 7
• 2 2
• y (2) - 1cos4 7 3.826 or 3.827
• 2 2
• d.) distance the integral from 0 to 2 of the
• absolute value of v(t) 1.173

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Works Cited

• Mr. Watson Class Notes and Worksheets
• Average velocity. Chapter 20-4 Velocity and
  Acceleration. 7 June 2007 lthttp//home.alltel.n
  et/?okrebs/?page205.htmlgt.
• Foerster, Paul A. Calculus. EmeryVille Steve
  Rasmussen, 1998.
• Problem of the Year 2003-2004. Calculus-Help.
  7 June 2007 lthttp//www.calculus-help.com/gt.