Physics, asked by aayushmittalktl2003, 1 year ago

A train travelling along a straight track with a speed
of 2 ms-1. begins to accelerate with
a = 2/v ms-2, where vis in ms 1. Its velocity after
3 seconds is
(1) 1ms-1
(2) 2 ms-1
(3) 3ms-1
(4) 4ms-1

Answers

Answered by abhi178
18

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 .

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