5ms+16=21ms initial velocity (io)
the delta v(t)=(ending velocity-io)/total deceleration time = (t)f/s^2
Thus, change in velocity = a t
0-21 = (t)f/s^2
second to top = t
and therefore:
d = Vi t + (1/2) a t^2
a particle is moving in a straight line the particle starts with speed 5ms and accelerates at a constant rate of 2ms for 8s it then decelerates at a constant rate coming to rest in a further 12s
find the total distance during 20s
it is 230 but i don't understand how
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