The instantaneous velocity of a particle moving in a straight line is given as V = α t + β t², where α and β are constants. The distance travelled by the particle between 1s and 2s is :

Solution

Physics sir · 4 steps

Velocity changes with time, so do not use s = vt. Integrate V from 1 s to 2 s.

1. Why integrate. The acceleration is not constant here, so the equations of uniform motion do not apply. Since V = dx/dt, the displacement over an interval is the integral of V over that interval.

   Δ x = ∫₁² (α t + β t²) dt = [(α t²)/2 + (β t³)/3]₁²

2. Put in limits. The α term gives α/2(4 - 1) = 3/2α. The β term gives β/3(8 - 1) = 7/3β.
3. Distance equals displacement. Taking α and β positive, V stays positive between 1 s and 2 s. The particle never turns back, so the distance equals the displacement.
4. So. The distance travelled is 3/2α + 7/3β, which is option (b).

The idea to keep. When velocity is a function of time, distance comes from integrating v dt between the two instants. Check that v does not change sign, otherwise split the integral at the turning point.

Answer from the NTA final answer key. Solution by PrepWiser.