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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)bt 2 b) ut+(1/2)bt 2 c) ut+(1/6)bt 3 d) ut+(1/3)bt 3

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)bt2


b) ut+(1/2)bt2


c) ut+(1/6)bt3


d) ut+(1/3)bt3


 

Grade:11

2 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

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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Mukunda Nagre
21 Points
3 years ago
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

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