Question

A traain is crossing an observer standing on platform. the first compartment of the train takes 2 second to cross the observer while the second compartment takes 2.5 seconds to cross him. the train is moving with uniform acceleration. find: (a) the velocity of the train when the front of the first compartment crosses the observer, and (b) the acceleration of the train. given, the length of each compartment is 15 metres.

Answer 1

a) When the front of the first compartment crosses the observer

then, distance covered = 15m (length of compartment)

time taken = 2s

Therefore velocity = 15/2 = 7.5 m/s

b) Velocity when second compartment crosses the observer =

15/2.5 = 6 m/s

So, acceleration is given by

a = change in speed / time

a = (6 - 7) / 0.5 = -3 m/s^2

Acceleration = -3 m/s^2

Answer 2

Hello Dear.

Here is the answer---

(a) For the First Compartments,
 Time taken(t) = 2 seconds.
Length of the Each Compartments = Distance
 = 15 m.

Thus, Speed of the Train(u) = 15/2
= 7.5 m/s.

∴ Velocity of the Train is 7.5 m/s.


(b) For the Second Compartments,
 Distance = Length of the Compartments 
= 15 m.
Time = 2.5 seconds.

∴ Speed(v) = 15/2.5
    = 6 m/s.


For the Acceleration,
   
Final Velocity(v) = 6 m/s.
Initial Velocity(u) = 7.5 m/s.
Time = 2.5 - 2
= 2 seconds.

     $Acceleration = \frac{v - u}{t}$
⇒ Acceleration = $\frac{6 - 7.5}{0.5}$
   = -1.5/0.5
  = - 3 m/s²

∴ Acceleration Produced by the Train is 3 m/s².


Hope it helps.