httpiimgurcommNp4x4qjpgSolutionFor a body under acceleration

http://i.imgur.com/mNp4x4q.jpg

Solution

For a body under acceleration, we know that, vdv/ds = acceleration = 32.2 - 50s

Hence vdv = (32.2 - 50s) ds

Integrating both the sides, until the point it stops

Hence 0 = 32.2S - 25S^2

That is, S = 32.2/25 = 1.288 metres

b.) Further, the velocity will be maximum when the acceleration becomes zero, thereaftere,

deceleration will start.

Integrating, the above equation, we get v^2/2 = 32.2S - 25S^2

At S = 32.2/50, acceleration becomes zero. Hence the maximum velocity will be obtained by

substituting this value of S in the above equation.

That is, velocity = sqrt(2*[32.2^2 / 50 - (32.2^2)*25 / 50^2]) = sqrt(20.7368) = 4.55 m/s