Question

Rest and Motion '; Kinematics'

A train starts from rest and moves with a constant acc of 2m/s2 for a half min. The brakes r then applied the train comes to rest in one min.Find the position(s )of d train at half the max speed.

Answer 1

The distance covered by the train in the first half minute,
$$\begin{gathered} s_1=u_1{t}_1+\frac{1}{2}{a}_1{t_1}^2=0+\frac{1}{2}\times 2\times {30}^2=900\text{m} \\ Thevelocityafter30s, \\ {u}_2=u_1+a_1{t}_1=0+2\times 30=60\text{m/s} \\ Letusfindtheaccelerationinsecondhalfminutebyu\sin gequationofmotion, \\ {v}_2=u_2+a_2{t}_2 \\ \Rightarrow 0=60+a_2\times 30 \\ \Rightarrow a_2=-2{\text{m/s}}^2 \\ Themaximumspeedis60m/s \\ Therefore,thepositionofthetrainat30m/s \\ {v_3}^2={u_1}^2+2a_1{s}_3 \\ \Rightarrow {30}^2=0+2\times 2\times s_3 \\ \Rightarrow s_3=900/4=225\text{m} \\ Letusfindthepositionofthetrainwhenthetrainisdeceleratingandreachesspeed30m/s \\ {v_4}^2={u_2}^2+2a_2{s}_4 \\ \Rightarrow {30}^2={60}^2-2\times 2\times s_4 \\ \Rightarrow s_4=\frac{{60}^2-{30}^2}{4}=675\text{m} \\ Totaldis\tan cecovered=s_1+s_4=900+675=1575\text{m} \\ Thereforeatdis\tan ces225\text{m}and1575\text{m},thetrainhashalfthemaximumspeed. \end{gathered}$$