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
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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,
or,
given, initial velocity of train , u = 2m/s
and we have to find velocity after time t = 3s
or,
or,
or,
or, ⇒v = 4m/s
hence, option (4) is correct choice .
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