Solve $0.1t^4 – 3.2t^{3/2}$ for particle coming to instantaneous rest

139 Views Asked by At

I need to factorize this to find real roots of "t"!! Question is about differentiation and integration

Q. A particle moves in a straight line. It starts from rest at a fixed point $O$ on the line. Its acceleration at time $t$ s after leaving $O$ is $a \,{\rm ms^{−2}}$, where $a = 0.4t^3 − 4.8t^{1/2}$!!! Show that, in the subsequent motion, the acceleration of the particle when it comes to instantaneous rest is $16 \,{\rm ms^{−2}}.$

For this I need to integrate equation of acceleration and find to values of "$t$" for which the integrated equation is equal to zero. What are values of "$t$" for which equation of velocity is equal to zero????

1

There are 1 best solutions below

0
On

Integrate to get the velocity and set it zero (rest)

$$v(t^*)=\int_0^{t^*} a(t)dt= 0.1(t^*)^4-3.2(t^*)^{3/2}=0$$

which factorizes as

$$0.1(t^*)^{3/2}[(t^*)^{5/2}-32]=0$$

and has the solution $(t^*)^{5/2}=32$, or $t^*=4s$.

Thus, the acceleration of the particle at the instantaneous rest time $t^*$ is $a(4)=16m/s^2$.