Answer to Question #303599 in Quantum Mechanics for qqq

Question

The position of a stone dropped from a cliff is described approximately by x = 5t^2 , where x is in meters and t is in seconds. The +x direction is downwards and the origin is at the top of the cliff. Find the velocity of the stone during its fall as a function of time t.

Answer 1

We know that gives

$x=5t^2$

Now velocity

$v=\frac{dx}{dt}$

Put value

$v=\frac{d(5t^2)}{dt}=10tm/sec$