Question

a carstarts fro rest and moves with unifrm accerleraton for 40s.it then moves with uniform velocity attaine by it and is finally brought to rest n 30s under uniform retaration.the car covers a total distance of 1380m n 2.5 mins .calculate constant speed,acceleration and retardation?

Answer 1

Let initial speed = u.
$$\begin{gathered} After40seconds, \\ v=a\left(40\right) \\ {s}_1=1/2a(40{)}^2=800a \\ Itcontinuestomoveatthisspeedforthenext80seconds. \\ {s}_2=40a\times 80=3200a \\ Next30seconds, \\ u=40a \\ v=0 \\ v=u-dt \\ 0=40a-d(30) \\ d=deceleration=-4/3a \\ {s}_3=40a(30)-1/2(4/3a\left)\right(30{)}^2 \\ =1200a-600a \\ =600a \\ So,totaldis\tan ce=800a+3200a+600a=1380 \\ a=0.3m/s^2 \\ d=-4/3\times 0.3=0.4m/s^2 \\ Cons\tan tspeed=40a=40*0.3=12m/s \end{gathered}$$