Starting from rest a vehicle accelerates at the rate of 2 m/s² towards east for 10 s. It then stops suddenly. It then accelerates again at a rate of 4√2 m/s² for next 10 s towards south and then again comes to rest. The net displacement of the vehicle from the starting point is

CONCEPT:

Kinematics and Vector Addition

• The displacement of an object is a vector quantity that has both magnitude and direction.
• When a vehicle accelerates from rest, the distance traveled can be calculated using the kinematic equation:

  s = ut + (1/2)at²

• For vector addition, the net displacement can be found by combining the displacements in each direction using the Pythagorean theorem.

EXPLANATION:

• First phase:

  Acceleration = 2 m/s² towards east

  Time = 10 s

  Initial velocity (u) = 0 (starting from rest)

  Distance traveled towards east (s_east) = ut + (1/2)at²

  • = 0 + (1/2) * 2 m/s² * (10 s)²
  • = (1/2) * 2 * 100
  • = 100 meters
• Second phase:

  Acceleration = 4√2 m/s² towards south

  Time = 10 s

  Distance traveled towards south (s_south) = ut + (1/2)at²

  • = 0 + (1/2) * 4√2 m/s² * (10 s)²
  • = (1/2) * 4√2 * 100
  • = 200√2 meters
• Net displacement:

  Use the Pythagorean theorem to find the resultant displacement (d):

  • d = √(s_east² + s_south²)
  • = √(100² + (200√2)²)
  • = √(10000 + 80000)
  • = √90000
  • = 300 meters

Therefore, the net displacement of the vehicle from the starting point is 300 meters.