A train moving with a constant acceleration crosses an observer standing on the platform. The first and the second compartments, each 15 m long cross the observer in 2s and 2.5 s, respectively. Find the velocity of the train when its first compartment just crosses the observer and also find its acceleration.

Explanation

Let the velocity and the acceleration of the train as its 1st compartment just reaches the observer be u and a respectively.
Hence, displacement in 2 s = length of the 1st compartment = 15 m and displacement in (2+2.5) or, 4.5s = total length of the two compartments = `2xx15 = 30 `m
Now from the equation s = `ut+(1)/(2)at^(2)` we get,
`:. " " 15 = 2u+(1)/(2)a(2)^(2)`
or, `" " u+a = (15)/(2)`
and ` 30 = 4.5 +(1)/(2)a(4.5)^(2)`
or, 36u+81a=240
Solving equation (1) and (2) we get
`u = (49)/(6) "and" a = -(2)/(3)`
The 1st compartment crosses the observer in 2 s, the velocity at that moment,
`v = u+at = (49)/(6)-(2)/(3)xx2 = (41)/(6)`
Therefore the velocity and acceleration of the train as its 1st compartment just crosses the observer are `(41)/(6) "m.s"^(-1)` and `-(2)/(3) "m.s"^(-2)` respectively.