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  • The acceleration of a particle increases with time “t” as   From the initial positio

The acceleration of a particle increases with time “t” as  at^{2}; From the initial position with a velocity u; the distance traveled by the particle in time “t” is : 

Option: 1

\frac{at^{4}}{12}+ut


Option: 2

\frac{at^{3}}{12}+t


Option: 3

\frac{at^{2}}{12}+ut


Option: 4

\frac{at}{12}+ut


Answers (1)

best_answer

From this question, we can say that 

Acceleration = \frac{dv}{dt}=at^{2}

And by integrating this equation we will get   v=\frac{at^{3}}{3}+C_{1}

 

At initial position t = 0 and v=u; therefore  C_{1}=u

Now we know velocity = \frac{dx}{dt}=\frac{at^{3}}{3}+u

After integrating this we will get displacement as                           x=\frac{at^{4}}{12}+ut+C_{2}

At t=0 , x=0 ; therefore  C_{2}=0

So we will get the distance as    \frac{at^{4}}{12}+ut


 

Posted by

avinash.dongre

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