InterviewSolution
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InterviewSolution
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Position vector `vec(r)` of a particle varies with time t accordin to the law `vec(r)=(1/2 t^(2))hat(i)-(4/3t^(1.5))hat(j)+(2t)hat(k)`, where r is in meters and t is in seconds. (a) Find suitable expression for its velocity and acceleration as function of time (b) Find magnitude of its displacement and distance traveled in the time interval `t=0` to `4 s`. |
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Answer» (a) Velocity `vec(v)` is defined as the first derivative vector with respect to time. `vec(v)=(dvec(r))/(dt)=t hat(i)-2sqrt(t)hat(j)+2hat(k) m//s` Acceleration `vec(a)` is defined as the first derivative of velocity vector with respect to time. `vec(a)=(dvec(v))/(dt)=hat(i)-1/sqrt(t)hat(j) m//s^(2)` (b) Displacement `Deltavec(r)` is defined as the change in place of position vector. `Deltavec(r)=8hat(i)-32/3hat(j)+8hat(k) m` Magnitude of displacement `Deltar=sqrt(8^(2)+(32/3)^(2)+8^(2))=1.55 m` Distance `Deltas` is defined as the path length and can be calculated by integrating speed over the concerned time interval. `Deltas=underset(0)overset(4)(int)vdt=underset(0)overset(4)(int) sqrt(t^(2)+4t+4)dt=underset(0)overset(4)(int)(t+2)dt=16 m` |
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