Lab 3, KINS 382

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Lab 3, KINS - 382

• Linear Acceleration

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Linear Acceleration

• Acceleration is a change in motion state.
• The symbol ais used to represent linear
  acceleration.
• Acceleration is measured in m/s2.

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The Equation

• a V t
• V a(t)
• t V a
• All of the above are derivatives of the same
  equation

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Question 1

• In her last sprint before retiring, sprinter Gail
  Devers moves at 4.34m/s at t 2sec and at t
  5sec she is moving at 9.12m/s.
• What is her average acceleration between her 2nd
  second and her 5th second?

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Answer 1

• a Vf - Vo/tf - to Vf 9.12m/s, Vo
  4.34m/s, tf 5sec, to 2sec
• (Step1) a 9.12m/s - 4.34m/s / 5sec - 2sec
• (Step2) a 4.78m/s / 3s 1.59m/s2

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Question 2

• An object is observed to have a velocity of 13m/s
  at t 2sec and a velocity of 19m/s at t 8sec.
• What is the average acceleration of the object?

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Answer 2

• a Vf - Vo / tf - to Vf 19m/s,
  Vo 13m/s, tf 8sec, to 2sec, a ?
• (Step1) a 19m/s - 13m/s / 8sec - 2sec
• (Step2) a 6m/s / 6s 1m/s2

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Question 3

• During the 100m dash Maurice Green has an average
  acceleration of 3.85m/s2 between 2sec and 6sec.
• What was his velocity during this time, in mi/hr?

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Answer 3

• a(tf - to) V a 3.85m/s2, tf 6sec, to
  2sec, V ?
• (Step1) V 3.85m/s2(6sec - 2sec)
• (Step2) V 3.85m/s2(4sec)
• (Step3) V 15.4m/s
• (Step4) 15.4m/s X 1mph/.447m/s 34.45mph
• This is probably not realistic!

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Question 4

• If the change in velocity of a racing dog is
  12.35 m/s, and the average acceleration is
  3.60m/s2.
• Calculate the change in time in seconds.

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Answer 4

• V a t V 12.35m/s, a 3.60m/s2, t ?
• (Step1) t 12.35m/s 3.60m/s2 3.43s
• Note The m/s cancel out leaving one of the
  seconds from the value representing acceleration.

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Question 5

• During a race a sprinter increases his velocity
  from 11ft/s at t 2sec (elapsed time) to 31ft/s
  at t 7sec.
• What was his average acceleration in m/s2?

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Answer 5

• Solve for a, Vf 31ft/s, Vo 11ft/s, tf 7sec,
  to 2sec, a ?
• (Step1) Convert ft/s to m/s 31ft/s X .3048m/s /
  1ft/s 9.45m/s 11ft/s X .3048m/s / 1ft/s
  3.35m/s
• (Step2) a 9.45m/s - 3.35m/s 7sec - 2sec
• (Step3) a 6.10m/s 5sec 1.22m/s2

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Question 6

• A cheetah has an average acceleration of 11m/s2.
  
• If the change in time is 2.6sec, find its change
  in velocity in mi/hr and m/s.

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Answer 6

• V a(t) a 11m/s2, t 2.6sec, V ?
• (Step1) V 11m/s2(2.6sec) 28.6m/s
• (Step2) 28.6m/s X 1mi/hr .447m/s 63.98mi/hr

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Question 7

• A discuss leaves the throwers hand with an
  initial velocity of 0m/s, and 5sec later, it has
  a velocity of 12.46m/s.
• What is the average acceleration of the discuss?

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Answer 7

• Solve for a Vf 12.46m/s, Vo 0m/s, tf
  5sec, to 0sec
• (Step1) a 12.46m/s - 0m/s 5sec - 0sec
  2.49m/s2

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Question 8

• Stanford Universitys Logan Tom kills a
  volleyball with a change in velocity of 58m/s.
• The average acceleration is 290m/s2
• Calculate the change in time and the velocity of
  the ball in mi/hr.

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Answer 8

• t V a V 58m/s, a 290m/s2,t ?
• (Step1) t 58m/s 290m/s2 .20sec
• (Step2) 58m/s X 1mi/hr .447m/s 129.75mi/hr

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Have a nice day.