A vehicle starts moving along a straight line path from rest. In first t seconds it moves with an acceleration of 2 m/s² and then in next 10 seconds it moves with an acceleration of 5 m/s². The total distance travelled by the vehicle is 550 m. The value of time t is

CONCEPT:

Equations of Motion with Constant Acceleration

• The equations of motion for an object moving with constant acceleration are:
  • v = u + at
  • s = ut + (1/2)at²
  • v² = u² + 2as
• Where:
  • u = initial velocity
  • v = final velocity
  • a = acceleration
  • t = time
  • s = distance travelled

EXPLANATION:

• In the given problem:
  • Initial velocity, u = 0 (since the vehicle starts from rest)
  • First acceleration, a₁ = 2 m/s²
  • Time for first acceleration, t₁ = t seconds
  • Second acceleration, a₂ = 5 m/s²
  • Time for second acceleration, t₂ = 10 seconds
• Distance travelled during the first acceleration (s₁):
  • s₁ = ut₁ + (1/2)a₁t₁²
  • = 0 + (1/2) * 2 * t₁²
  • = t₁² meters
• Velocity at the end of first acceleration (v₁):
  • v₁ = u + a₁t₁
  • = 0 + 2t₁
  • = 2t₁ m/s
• Distance travelled during the second acceleration (s₂):
  • s₂ = v₁t₂ + (1/2)a₂t₂²
  • = (2t₁) * 10 + (1/2) * 5 * 10²
  • = 20t₁ + 250 meters
• Total distance travelled (s_total):
  • s_total = s₁ + s₂
  • = t₁² + 20t₁ + 250
  • Given s_total = 550 meters
  • Therefore, t₁² + 20t₁ + 250 = 550
  • t₁² + 20t₁ - 300 = 0

Solving the quadratic equation t₁² + 20t₁ - 300 = 0:

• (t₁ + 30)(t₁ - 10) = 0
• t₁ = -30 (not physically possible) or t₁ = 10

Therefore, the value of time t is 10 seconds.