Question

Two stations A and B are 10 km apart in a straight track, and a train starts from A and comes to rest at B. For three quarters of the distance, the train is uniformly accelerated and for the remainder distance uniformly retarded. If it takes 15 minutes over the whole journey, find its acceleration, retardation and the maximum speed it attains.

Answer 1

For three quarters, the acceleration $=a_1(forward)$

For the remaining quarter, acceleration $=-a_2$

Time taken to cover three quarter's distance $=t_1$

Time taken to cover remaining quarter $=t_2$

Total time taken, $t_1+t_2=(15\times60)=900\space s$

Initial velocity of train when it starts from A, $u=0$

$s_1=ut^2_1+\dfrac{1}{2}a_1t_1^2$

$\dfrac{3}{4}\times(10\times10^3)=\dfrac{1}{2}a_1t_1^2$

$t_1=\sqrt{\dfrac{3\times10^4}{2a_1}}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(1)$

$v^2-u^2=2as_1$

$v=\sqrt{2\times a_1\times\dfrac{3\times10^4}{4}}=\sqrt{\dfrac{3\times10^4a_1}{2}}$ $\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(2)$

For remaining quarter,

initial velocity = $v$

final velocity =

$0-v^2=2(-a_2)s_2$

$v=\sqrt{2\times a_2\times\dfrac{1\times10^4}{4}}=\sqrt{\dfrac{10^4a_2}{2}}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(3)$

$0=v+(-a_2)t_2$

$t_2=\sqrt{\dfrac{10^4a_2}{2}}\times\dfrac{1}{a_2}$

$t_2=\sqrt{\dfrac{10^4}{2a_2}}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(4)$

From equation (2) and (3)

$3a_1-a_2=0$

$a_2=3a_1\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(5)$

Adding equation (1) and (4)

$t_1+t_2=\sqrt{\dfrac{3\times10^4}{2a_1}}+\sqrt{\dfrac{10^4}{2a_2}}$

$\sqrt{\dfrac{3\times10^4}{2a_1}}+\sqrt{\dfrac{10^4}{2a_2}}=900$

$\sqrt{\dfrac{3}{2a_1}}+\sqrt{\dfrac{1}{6a_1}}=9$

$a_1=0.0329\space m/s^2$

$a_2=0.0987 \space m/s^2$

$v=22.214\space m/s$