Answer to Question #78004 in Mechanics | Relativity for Khyu

Question

A train in straight line motion passes a point A with a velocity of 6 ms^-1. It moves from A to B with a constant acceleration a1 ms^-2 and reaches B in 10s. From B to C it has a constant acceleration a2. It takes 4s to move from B to C. Find a relationship between a1 and a2 if the velocity at C is 54ms^-1.

Answer 1

Solution:
speed at point B is Vb=Va+A1*t1
speed at point C is Vc=Vb+A2*t2=Va+A1*t1+A2*t2
Va is 6 m/s
Vc is 54 m/s
So
54=6+a1*10+a2*4
48=a1*10+a2*4
24=a1*5+a2*2
So
a2=(24-a1*5)/2