InterviewSolution
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InterviewSolution
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A particle is moving in a straight line. Its displacement at any instant t is given by `x = 10 t+ 15 t^(3)`, where x is in meters and t is in seconds. Find (i) the average acceleration in the intervasl t = 0 to t = 2s and (ii) instantaneous acceleration at t = 2 s. |
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Answer» The given equation, `x = 10 t + 15 t^(3)` and the variables are (i) t = 0 to t = 2 s (ii) t = 2 s We need to clarify all the variables. Velocity of particle, `v=(dx)/(dt)` While calculating instantaneous velocity , we need to differentiate. Differentiating x w.r.t. time. `v=(d)/(dt)(10 t+15t^(3))=10+45 t^(2)` At `t = 0, V_(0) = 10+45(0)=10 ms^(-1)` At t= 2 s, `V_(2)=10+45xx(2)^(2)=10+180=190 ms^(-1)` `Delta V= V_(2)-V_(1)=190-10=180 ms^(-1)` `Delta t=2-0=2s` `therefore a_(av)=(Delta V)/(Delta t)=(180)/(2)=90 ms^(-2)` Now, differentiate, v with respect to time ti final the acceleration of the particle. `a=(d)/(dt)(10+45 t^(2))=90 t` At `t=2s, a = 90xx2 = 180 ms^(-2)` The instantaneous acceleration of a particle at 2 is `180 ms^(-2)`. |
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