Question

A car starting from rest, accelerates at the rate a through a distance s, then continues at constant speed for time t and then decelerates at the rate a/2 to come to rest. If total distance is traversed in 15 sec f . Find s​

Answer 1

Answer:

S = \frac{1}-{2} [u+v]t

              v = u+at

            v-u = at

                t = \frac{v-u}-{a}

                S = \frac{1}-{2} [u+v]t

                   = \frac{1}-{2} [u+v][\frac{u-v}-{a} ]

              as = \frac{1}-{2} [u+v][u-v]

            2as = [u+v][u-v]

                   = v^{2} - u^{2}

Explanation: