Linear Motion

19/10/2010

Terminology and SI (Standard International) Units

Distance metres (m)
Displacement metres (m)
Speed metres per second (m/s or ms-1)
Velocity metres per second (m/s or ms-1)
Acceleration metres per second per second (m/s/s, m/s2 or ms-2)
Time seconds (s)

Speed and Velocity

Speed = distance m/s or ms-1
time

Displacement-Time graphs

  • Displacement (distance) goes on the vertical axis
  • Time goes along the horizontal axis
  • The gradient of the line is:
    change in displacement = velocity
    time taken
  • A straight line indicates constant velocity
  • A curved line indicates a change in velocity so the object is accelerating
  • The steeper the gradient, the greater the velocity

Using the graph

the total displacement = average velocity
the total time
the total distance = average speed
the total time
  • Horizontal lines mean no change in displacement, i.e. the object is stationary
  • A positive slope means the object is moving away from its starting point
  • A negative slope means the object is moving closer to its starting point

Velocity-Time graphs

  • Velocity goes on the vertical axis
  • Time goes along the horizontal axis
  • The gradient of the line is:
    change in velocity = acceleration
    time taken
  • A straight line indicates constant (uniform) acceleration
  • A curved line indicates a non-uniform acceleration
  • The steeper the gradient, the greater the acceleration

Using the graph

  • Horizontal lines mean no change in velocity, i.e. the object is not accelerating
  • A positive slope means the object is accelerating
  • A negative slope means the object is decelerating
  • The area under the graph gives the displacement (distance)
  • If the graph goes below the horizontal axis it shows motion in the opposite direction

"S.U.V.A.T" Mnemonic

Displacement: s
Initial velocity: u
Final velocity: v
Acceleration: a
Time: t

Useful equations

These equations can only be used if the acceleration is constant
s = ½(u + v)t
s = ut + ½at2
If 'u' is not given, it is most often assumed to be 0ms-1
v = u + at
v2 = u2 + 2as

Click here to show a further 14 rearranged versions of the above equations

Rearranged Equations

s = v2 - u2
2a
s = vt - ½at2
u = v - at
u2 = v2 - 2as
u = 2s - v
t
u = s - ½at
t
v = u + at
v2 = u2 + 2as
v = 2s - u
t
v = s + ½at
t
a = v2 - u2
2s
a = v - u
t
a = 2 ( s - ut )
t2
a = 2 ( vt - s )
t2
t = 2s
u + v
t = v - u
a

Useful text on rearranging equations
http://www.andy-roberts.net/teaching/ma11/transposing.pdf

Questions

  1. A cheetah starts from rest and accelerates at 2.0ms-2 due east for 10s. Calculate:
    1. the cheetah's final velocity
      • s = ?
      • u = 0ms-1
      • v = ?
      • a = 2ms-2
      • t = 10s
      • v = u + at
      • v = 0 + (2 x 10)
      • v = 20

      Answer: final velocity = 20ms-1
    2. the distance the cheetah covers in this 10s
      • s = ?
      • u = 0ms-1
      • v = 20ms-1
      • a = 2ms-2
      • t = 10s
      Always use the formula that uses the most given values.
      • s = ut + ½at2
      • s = (0 x 10) + (½ x 2 x 100)
      • s = 100

      Answer: distance covered = 100m


  2. An athlete accelerates out of her blocks at 5.0ms-2.
    1. How long does it take her to run the first 10m?
      • s = 10m
      • u = 0ms-1
      • v = ?
      • a = 5ms-2
      • t = ?
      • First work out v using v2 = u2 + 2as
      • v2 = 0 + (2 x 5 x 10)
      • v2 = 100
      • v = 10
      • Then t = v - u
        a
      • t = 2

      Answer: Time taken = 2s
    2. What is her velocity at this point?
      • s = 10m
      • u = 0ms-1
      • v = ?
      • a = 5ms-2
      • t = 2s
      • v2 = u2 + 2as
      • v2 = 0 + (2 x 5 x 10)
      • v2 = 100
      • v = 10

      Answer: Final velocity = 10ms-1


  3. A bicycle's brakes can produce a deceleration of 2.5ms-2.

    How far will the bicycle travel before stopping, if it is moving at 10ms-1 when the brakes are applied?

    Converting the given deceleration value to an acceleration value of -2.5ms-2 will be worth a mark in an exam.
    • s = ?
    • u = 10ms-1
    • v = 0ms-1
    • a = -2.5ms-2
    • t = ?
    • Rearrange v2 = u2 + 2as to solve for s
    • s = v2 - u2
      2a
    • s = -100
      -5
    • s = 20

    Answer: Stopping distance = 20m

Physics Books in the Library

You can find these on the library's website by searching the ISBN.

  • Introductory Physics by Jerold Touger
  • ISBN 0471940003
  • Effective Communication for Science and Technology
  • (Palgrave Study Guide)
  • ISBN 0333775465
  • The Sciences Good Study Guide
  • ISBN 0749234113

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