A maths

• starmoonsun

  19 Jul 07, 16:13

  1) A car starts from rest and accelerates uniformly at a ms-2 fot t1 seconds to a speed of 10ms-1. it then accelerates at 2a ms-2 for a futher t2 seconds to a speed of 20ms-1. it maintains this speed for a further 24 seconds. the total distance covered is 555m.
  a) show that t1 = t2.
  c) the total time taken
  d) the value of a
  
  2) a motorcyclist travelling along a str road passes O with a speed of 10m/s and cont'd at this speed for t1 seconds. over the next 10 seconds he accelerates at a constant rate to a speed of 15m/s. he then brings the motorcycle to rest in a further t2 seconds by retarding at a constant rate. his acceleration and retardation are of equal magnitude.
  
  given the total distance is 850m, calculate t1
  
  anyone can help me? thanks!
• eagle

  19 Jul 07, 17:17

  Originally posted by starmoonsun:

  1) A car starts from rest and accelerates uniformly at a ms-2 fot t1 seconds to a speed of 10ms-1. it then accelerates at 2a ms-2 for a futher t2 seconds to a speed of 20ms-1. it maintains this speed for a further 24 seconds. the total distance covered is 555m.
  a) show that t1 = t2.
  c) the total time taken
  d) the value of a
  

  a)
  
  t1 = (10-0) / a = 10 / a
  
  t2 = (20 - 10) / 2a = 5 / a
  
  So t1 = 2* t2. Please check again
  
  b)
  use s = 1/2 (v+u) t
  v = final velocity
  u = initial velocity
  s = displacement
  
  first part of travel: s = 1/2 * 10 * t1 = 10 * t2 since t1 = 2 * t2
  2nd part of travel: s = 1/2 * 15 * t2 = 7.5 * t2
  3rd part of travel: s = 24s * 20m/s = 480m
  
  hence, 17.5 * t2 = 555 - 480 = 75
  t2 = 4.29 s
  total time = t1 + t2 + 24 = 36.9s
  
  c) 5/a = 4.29
  a = 1.17 m / s^2
  
  
• eagle

  19 Jul 07, 17:24

  Originally posted by starmoonsun:

  2) a motorcyclist travelling along a str road passes O with a speed of 10m/s and cont'd at this speed for t1 seconds. over the next 10 seconds he accelerates at a constant rate to a speed of 15m/s. he then brings the motorcycle to rest in a further t2 seconds by retarding at a constant rate. his acceleration and retardation are of equal magnitude.
  
  given the total distance is 850m, calculate t1
  

  Constant Speed part:
  Dist travelled = 10 * t1
  
  Acceleration part:
  Using s = 1/2 (v+u) t,
  v = final vel = 15
  u = initial vel = 10
  Dist travelled = 1/2 * (10+15) * 10 = 125m
  
  Deceleration part:
  acceleration is (15 - 10) / 10 = 0.5 m /s^2
  Hence deceleration = - 0.5 m / s^2
  
  Using v = u + at, where v = 0, u = 15, a = -0.5, t = t2
  t = 30s
  Dist travelled = 1/2 * (15 + 0) * 30 = 225m
  
  Hence,
  10 * t1 + 125 + 225 = 850
  Solving gives t1 = 50s