Kinematics: compare time and speed

Question

How to solve this problem:

The first train half of its time moved with 80 km/h, another half with 40 km/h. The second train went slowly accelerating backward from 80 km/h to 40 km/h. Which train took more time moving and which train's halfway velocity was greater.

The correct answer: t1=t2, v1>v2

Answer 1

At first we need to make some clarifications: the motion is uniform accelerated and distance for both trains is the same. Than gives a=constant, d₁=d₂=d; let’s denote t₁ as time of the first train and t₂ as time of the second one.

 For the first train we can write d= (t₁/2)(80+40)=60 t₁, then t₁=d/60; as for its halfway velocity we can state that the velocity V₁=80 km/s, as during halftime the second train covers more than its halfway.

 For the second train, as it moves uniform accelerated d=V(0)t + at^2/2  where V(0)=80 km/h ,

a=(40-80)/t₂=-40/t₂; at the same time we can use other kinematic equation Vfin^2 – Vin^2=2aD for D=d or

40 ^2 – 80^2=2ad= 2d(40-80)/t₂ , and then after simplifying (40+80)=2d/t₂; hence t₂=2d/120=d/60

Therefore, t₂₌ t₁=t= d/60

Now, try to find V₂ at halfway. Again use kinematic equation

Vfin^2 – Vin^2=2aD for D=d/2, that is (V₂)^2 -80^2=2a(d/2)=ad and using a=-40/t₂ and t= d/60 that were found in previous steps, we can write (V₂)^2=80^2 -(40*60); hence, V₂<80, that is,  V₂< V₁