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  4. Motion along a straight line
  5. Finding velocity and displacement

3.6 Finding velocity and displacement from acceleration  (Page 2/6)

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  1. From the functional form of the acceleration we can solve [link] to get v ( t ):
    v ( t ) = ∫ a ( t ) d t + C 1 = ∫ − 1 4 t d t + C 1 = − 1 8 t 2 + C 1 .

    At t = 0 we have v (0) = 5.0 m/s = 0 + C 1 , so C 1 = 5.0 m/s or v ( t ) = 5.0 m/ s − 1 8 t 2 .
  2. v ( t ) = 0 = 5.0 m/ s − 1 8 t 2 ⇒ t = 6.3 s
  3. Solve [link] :
    x ( t ) = ∫ v ( t ) d t + C 2 = ∫ ( 5.0 − 1 8 t 2 ) d t + C 2 = 5.0 t − 1 24 t 3 + C 2 .

    At t = 0, we set x (0) = 0 = x 0 , since we are only interested in the displacement from when the boat starts to decelerate. We have
    x ( 0 ) = 0 = C 2 .

    Therefore, the equation for the position is
    x ( t ) = 5.0 t − 1 24 t 3 .
  4. Since the initial position is taken to be zero, we only have to evaluate x ( t ) when the velocity is zero. This occurs at t = 6.3 s. Therefore, the displacement is
    x ( 6.3 ) = 5.0 ( 6.3 ) − 1 24 ( 6.3 ) 3 = 21.1 m .
Graph A is a plot of velocity in meters per second as a function of time in seconds. Velocity is five meters per second at the beginning and decreases to zero. Graph B is a plot of position in meters as a function of time in seconds. Position is zero at the beginning, increases reaching maximum between six and seven seconds, and then starts to decrease.
(a) Velocity of the motorboat as a function of time. The motorboat decreases its velocity to zero in 6.3 s. At times greater than this, velocity becomes negative—meaning, the boat is reversing direction. (b) Position of the motorboat as a function of time. At t = 6.3 s, the velocity is zero and the boat has stopped. At times greater than this, the velocity becomes negative—meaning, if the boat continues to move with the same acceleration, it reverses direction and heads back toward where it originated.

Significance

The acceleration function is linear in time so the integration involves simple polynomials. In [link] , we see that if we extend the solution beyond the point when the velocity is zero, the velocity becomes negative and the boat reverses direction. This tells us that solutions can give us information outside our immediate interest and we should be careful when interpreting them.

Check Your Understanding A particle starts from rest and has an acceleration function 5 − 10 t m/s 2 . (a) What is the velocity function? (b) What is the position function? (c) When is the velocity zero?

  1. The velocity function is the integral of the acceleration function plus a constant of integration. By [link] ,
    v ( t ) = ∫ a ( t ) d t + C 1 = ∫ ( 5 − 10 t ) d t + C 1 = 5 t − 5 t 2 + C 1 .
    Since v (0) = 0, we have C 1 = 0; so,
    v ( t ) = 5 t − 5 t 2 .
  2. By [link] ,
    x ( t ) = ∫ v ( t ) d t + C 2 = ∫ ( 5 t − 5 t 2 ) d t + C 2 = 5 2 t 2 − 5 3 t 3 + C 2 .
    Since x (0) = 0, we have C 2 = 0, and
    x ( t ) = 5 2 t 2 − 5 3 t 3 .
  3. The velocity can be written as v ( t ) = 5 t (1 – t ), which equals zero at t = 0, and t = 1 s.
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Summary

  • Integral calculus gives us a more complete formulation of kinematics.
  • If acceleration a ( t ) is known, we can use integral calculus to derive expressions for velocity v ( t ) and position x ( t ).
  • If acceleration is constant, the integral equations reduce to [link] and [link] for motion with constant acceleration.

Key equations

Displacement Δ x = x f − x i
Total displacement Δ x Total = ∑ Δ x i
Average velocity v – = Δ x Δ t = x 2 − x 1 t 2 − t 1
Instantaneous velocity v ( t ) = d x ( t ) d t
Average speed Average speed = s – = Total distance Elapsed time
Instantaneous speed Instantaneous speed = | v ( t ) |
Average acceleration a – = Δ v Δ t = v f − v 0 t f − t 0
Instantaneous acceleration a ( t ) = d v ( t ) d t
Position from average velocity x = x 0 + v – t
Average velocity v – = v 0 + v 2
Velocity from acceleration v = v 0 + a t ( constant a )
Position from velocity and acceleration x = x 0 + v 0 t + 1 2 a t 2 ( constant a )
Velocity from distance v 2 = v 0 2 + 2 a ( x − x 0 ) ( constant a )
Velocity of free fall v = v 0 − g t (positive upward)
Height of free fall y = y 0 + v 0 t − 1 2 g t 2
Velocity of free fall from height v 2 = v 0 2 − 2 g ( y − y 0 )
Velocity from acceleration v ( t ) = ∫ a ( t ) d t + C 1
Position from velocity x ( t ) = ∫ v ( t ) d t + C 2

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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