Question

A train begins at the origin and moves at constant speed v₁=15.2 m/s toward another train which is initially at rest at a distance d=164 m from the first train. If the second train has an acceleration of magnitude a₂=1.02 m/s² toward the first train, how far does the first train move before the trains collide? (Hint : use quadratic equation and choose positive time)

Answer 1

Time taken for the trains to collide = T

Constant speed of the first train = V₁ = 15.2 m/s

Distance traveled by the first train before the collision = D₁

D₁ = V₁T

Initial speed of the second train = V₂ = 0 m/s (At rest)

Acceleration of the second train = a₂ = 1.02 m/s²

Distance traveled by the second train before the collision = D₂

D₂ = V₂T + a₂T²/2

Initial distance between the two trains = D

The initial distance between the two trains is equal to the sum of the distances the two trains travel before colliding.

D = D₁ + D₂

D = V₁T + V₂T + a₂T²/2

164 = (15.2)T + (0)T + (1.02)T²/2

0.51T² + 15.2T - 164 = 0

$T = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

$T = \frac{-(15.2) \pm \sqrt{(15.2)^{2} - 4(0.51)(-164)}}{2(0.51)}$

$T = \frac{-15.2 \pm 23.78}{1.02}$

T = 8.412 sec or -38.2 sec

Time cannot be negative.

T = 8.412 sec

D₁ = V₁T

D₁ = (15.2)(8.412)

D₁ = 127.9 m

Distance the first train moves before the trains collide = 127.9 m