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Question

The motion of a body is given by the equation dv(t)dt=6.03v(t). Where v(t) is speed in ms and t in sec. If body was at rest at t = 0


A
The terminal speed is
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B
The speed varies with the time as
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C
The speed is when the acceleration is half the initial value
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D
The magnitude of the initial acceleration is
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Solution

The correct option is B The speed varies with the time as
dvdt=63vdv63v=dt
Integrating both sides, dv63v=dt
loge(63v)3=t+K1
loge(63v)=3t+K2               ......(i)
At t=0,v=0 loge6=K2
Substituting the value of K2 in equation (i)
loge(63v)=3t+loge6
loge(63v6)=3t e3t=63v6
v=2(1e3t)
vterminal=2ms(whent=)
Acceleration a=dvdt=ddt[2(1e3t)]=6e3t
Initial acceleration = 6ms2.

Physics

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