A particle moving along x−axis has acceleration f, at time t, given by f=f0(1−tT), where f0 and T are constants. The particle at t=0 has zero velocity. In the time interval between t=0 and the instant when f=0, the particle's velocity vx is:
f0T2
12f0T
f0T
12f0T2
A
12f0T
B
f0T
C
12f0T2
D
f0T2
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Solution
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let any time t particle has a velocity u and in a change in time (t+dt) its final velocity is u+dv, this dt time acceleration is constant
using first eq
v=u+at
u+du=u+f0(1−tT)dt
∫v0du=∫T0f0(1−tT)dt
v=f0(t−t22T)T0
v=f0(T−T2)
v=f0T2
Hence the option A correct
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