Question

a particle initially at rest moves along x-axis. Its acceleration is given as a = (2t+2) m/s2. If it starts from origin , find distance covered in 2 second.

Joshi sir comment

dv/dt = 2t+2

so dv = 2tdt+2dt

on integrating the given eq.

v = t2 + 2t + C

at start v = 0 at t = 0

so on puting these initial values we get C = 0 

so v = t2 + 2t 

so dx/dt = t2 + 2t 

so dx = t2dt + 2tdt 

on integrating x = t3/3 + t2+ C

on putting initial values x=0 at t=0 

we get C = 0

so x =  t3/3 + t

now put t = 2

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