Question

A cheetah accelerates at 13 m/s^2 on smooth ground in a straight line which forms an angle of 33 degrees (with x as a reference direction) after 2.3 seconds the velocity of the animal in that direction will be??

Answer 1

If the cheetah starts from rest and moves up along the inclined plane with net acceleration(acceleration due to his effort - component of gravitational acceleration acting downwards) of `13 m/s^2`,then after `2.3 s` its velocity will be `v = 0 + 13*2.3` (using `v=u+at`) so, `v = 29.9 m/s`

But if `13 m/s^2` is the maximum acceleration which it can provide due to its effort,then net acceleration upwards will be `(13-g sin 33) m/s^2` as a component of gravitational acceleration tries to oppose its motion ,in that case using the same equation velocity comes out to be `17.36 m/s`