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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)bt^2

b) ut+(1/2)bt^2

c) ut+(1/6)bt^3

d) ut+(1/3)bt^3

Grade:11

2 Answers

Shaswata Biswas
132 Points
7 years ago
Initial velocity = u. Aceeleration = bt. Time = t.
Then from equations of motion, displacement of the paricle, s = ut + \frac{1}{2}bt^{2}
So, option b is correct.
THANKS
Shaswata Biswas
132 Points
7 years ago
Sorry, I've made a mistake.
We have, aceeleration, a = bt
\implies \frac{\mathrm{dv} }{\mathrm{d} t} = bt
\implies dv = bt.dt
Integrating on both sides, 
\implies v = \frac{1}{2}bt^{2} + c
At t = 0, v = u. So, c = u
\implies v = \frac{1}{2}bt^{2} + u
\implies \frac{\mathrm{ds} }{\mathrm{d} t} = u + \frac{1}{2}bt^{2}
\implies ds = [u + \frac{1}{2}bt^{2} ]dt
Integrating both sides : 
\implies s = ut + \frac{1}{6}bt^{3} + c
At t = 0 the particle was at origin, so c = 0
Hence, distance travelled by the particle is  s = ut + \frac{1}{6}bt^{3}
Option c is correct
THANKS

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