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

The problem can be solved using the standard equations of motion in two phases:

• Phase 1: The vehicle starts from rest (u = 0) with acceleration a = 2 m/s² for time t. The distance covered is s₁ = ut + ½at² = 0 + ½(2)t² = t². The velocity at the end of this phase is v = u + at = 2t.
• Phase 2: For the next 10 seconds, the initial velocity is 2t and acceleration is 5 m/s². The distance covered is s₂ = ut + ½at² = (2t)(10) + ½(5)(10)² = 20t + 250.

The total distance is given as 550 m. Therefore:
t² + 20t + 250 = 550
t² + 20t - 300 = 0

Solving this quadratic equation: (t + 30)(t - 10) = 0. Since time cannot be negative, t = 10 seconds. Thus, Option A is correct.