Constant Acceleration

• Velocity-time graph


  • Cyclist moves at constant speed of 6 m/s for first 8 seconds, then decelerates at constant rate, stopping after 12 seconds
• Finding displacement from starting point after 12 seconds


  Displacement = area under the graph

  s = 1/2 (a + b)h

  s = 1/2 (8 + 12)6

  s = 60 m
• Finding deceleration rate


  Acceleration is the gradient of the slope

  a = (0 - 6) / (12 - 8) = -1.5 m/s^2
• Constant acceleration formulae 1


  s is displacement, u is initial velocity, v is final velocity, a is acceleration, t is time

  v = u + at

  s = (u + v)/2 * t
• Cyclist accelerating from 4 m/s to 7.5 m/s in 40 seconds

  • Distance travelled = (4 + 7.5)/2 * 40 = 230 m

  Acceleration = (7.5 - 4)/40 = 0.0875 m/s^2
• Constant acceleration formulae 2


  v^2 = u^2 + 2as

  s = ut + 1/2 at^2

  s = vt - 1/2 at^2
• Particle moving from A to B with constant acceleration 5 m/s^2


  • Distance from A to B = (18^2 - 3^2) / (2 * 5) = 31.5 m
• Vertical motion under gravity


  Acceleration due to gravity g = 9.8 m/s^2 (downward positive, upward negative)
• Displacement-time graphs


  Displacement on vertical axis, time on horizontal axis

  Gradient represents velocity
• Velocity-time graphs


  Velocity on vertical axis, time on horizontal axis

  Gradient represents acceleration

  Area under graph represents distance travelled