Question

v=4t-(1/16)t^3
Find the set of values of t for which the acceleration of particle is positive.

Answer 1

acceleration = dv/dt
find the range of t where dv/dt>0

Answer 2

Ok. So dv/dt= 4- (3/16)t^2
When 4-(3/16)t^2 > 0, the value of t>4.62.

Answer 3

But the answer is actually 0<t<4.62. How's that? :O

Answer 4

Ok, I think I found my mistake. I didnot change the inequality sign. This would be t<4.62. Thanks anyway!