A particle moves along a straight line with equation of motion s = t^{5} - 3 t^{4} Find the value of t (other than 0 ) at which the acceleration is equal to zero.

Well we have to resort to use of calculus in this question. Specifically, differentiation needs to be used.

s= (t^5) - {3(t^4)}

Now, velocity is the time-differential of distance. Differentiating the given equation with-respect-to-time, we get

v={5(t^4)} - {12(t^3)}

Again, acceleration is the time-differential of velocity. So differentiating the above equation with-respect-to-time, we get

a= {20(t^3)} - {36(t^2)}

Now, since we want to calculate those values of 't' at which 'a' is zero, we equate the right-hand-side of the above equation to 0.

{20(t^3)} - {36(t^2)} = 0

Taking 4 & (t^2) common from left-hand-side, we get

{4(t^2)} {(5*t) - (9)} = 0

Using basic concepts of mathematics, we know that this equation implies that

1) Either 't' = 0
(Which we've already been given in the question itself)

2) Or {(5*t) - (9)} = 0

From this,

t= (9/5) = 1.8 second

So the particle is acceleration will be zero at both t=0 & at t=1.8 second.

If you're not aware of calculus or differentiation, l suggest that u pick up a book that gives introductory ideas about it.

Hope that was helpful to u :)

Cheers :)

yeah. what he said :)