A train travelling along a straight track with a speedof 2 ms-1. begin

1.	A train travelling along a straight track with a speedof 2 ms-1. begins to accelerate witha = 2/v ms-2, where vis in ms 1. Its velocity after3 seconds is(1) 1ms-1(2) 2 ms-1(3) 3ms-1(4) 4ms-1
Answer» answer : OPTION (4) 4m/s explanation : Acceleration varies with velocity as , a = 2/v m/s² ......(1) we know, acceleration is the RATE of change of velocity. so, a = ∆v/∆t and instantaneous acceleration, a = dv/dt , use this APPLICATION in above expression (1) . a = dv/dt = 2/v or, $\int\limits^v_u{v}\,dv=2\int\limits^t_0{dt}$ or, $\left[\frac{v^2}{2}\right]^v_u=2t$ given, initial velocity of train , u = 2m/s and we have to find velocity after time t = 3s or, $\left[\frac{v^2}{2}\right]^v_2=2(3)$ or, $\frac{v^2-2^2}{2}=2(3)$ or, $v^2-4=12$ or, $v^2=16$ ⇒v = 4m/s hence, option (4) is correct choice .