Question

A girl rolls a basketball across a basketball court. The ball slowly decelerates at a rate of -0.20m/s^2. If the initial velocity was 2.0m/s and the ball rolled to a stop at 5.0sec after. 12:00, at what time did she start the ball rolling?​

Answer 1

By equation of motion :

$2as=v^2-u^2\\\\s=\dfrac{v^2-u^2}{2a}\\\\s=\dfrac{0^2-2^2}{2\times( -0.2)}\\\\s=10\ m$

Now, it is given that it stops after 5 seconds of motion and time at that point is 12:00.

So, time when it started is 11 : 59 : 55.

Therefore, distance travelled is 10 m.

Hence, this is the required solution.