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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)bt2
b) ut+(1/2)bt2
c) ut+(1/6)bt3
d) ut+(1/3)bt3
```
7 years ago

147 Points
```										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
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very  quickly. We are all IITians and here to help you in your IIT JEE  & AIEEE preparation. All the best. Regards,Askiitians ExpertsBadiuddin
```
7 years ago
Mukunda Nagre
21 Points
```										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
```
5 months ago
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