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
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
Initial velocity (u₁) = 0 m/s
Acceleration (a₁) = 2 m/s²
Time (t₁) = 10 s
Displacement in Phase 1 (s₁) = u₁t₁ + (1/2)a₁t₁² = 0*10 + (1/2)*2*(10)² = 100 m East. The vehicle stops suddenly after this displacement.
Phase 2 (towards South):
Starts from rest again, so initial velocity (u₂) = 0 m/s
Acceleration (a₂) = 4√2 m/s²
Time (t₂) = 10 s
Displacement in Phase 2 (s₂) = u₂t₂ + (1/2)a₂t₂² = 0*10 + (1/2)*(4√2)*(10)² = (1/2)*4√2*100 = 2√2*100 = 200√2 m South.
The net displacement is the vector sum of s₁ (100 m East) and s₂ (200√2 m South). These two displacements are perpendicular.
The magnitude of the net displacement (s_net) is found using the Pythagorean theorem:
s_net² = s₁² + s₂²
s_net² = (100)² + (200√2)²
s_net² = 10000 + (40000 * 2)
s_net² = 10000 + 80000
s_net² = 90000
s_net = sqrt(90000) = 300 m.
The direction of the net displacement would be South-East, but only the magnitude is asked.