The acceleration of a particle is increasing linearly with time t as bt. The particle starts from origin with initial velocity u. The distance travelled by the particle in time t will be:a) ut+(1/3)bt2b) ut+(1/2)bt2c) ut+(1/6)bt3d) ut+(1/3)bt3

Dear abhingya

d²s/dt² =bt

intigrate

V= ds/dt = bt²/2 +c

at t=0 V=U

so U =0 +c

ds/dt = bt²/2 +U

furthet intigrate

s= b/6 t³ +Ut +d

at t=0 s=0

so d=0

S =b/6 t³ +Ut

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Badiuddin

We have to assume acceration =0 when t=0. So we have the equation: a = bt Since a = dv/dt dv/dt = bt Integrating gives: v = (b/2)t² + C When t=0, v = u, so C = u v = (b/2)t² + uSince v = ds/dt ds/dt = (b/2)t² + uIntegrating gives: s = (b/6)t³ + u.t + K When t=0, s=0, so K=0 s= (b/6)t³ + u.t