Question

A train begins at the origin and moves at constant speed V1 =18.2 m/s toward another train which is initially at rest at a distance d=242 m from the first train. If the second train has an acceleration of magnitude az=0.99 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) Select one: 245.38 m 302.01 m 75.50 m 188.76 m 132.13 m

Answer 1

Initial distance between the two trains = d = 242 m

Time taken by the trains to collide with each other = t

Speed of the first train = v₁ = 18.2 m/s

Distance traveled by the first train before the collision = d₁

The first trains moves with a constant speed therefore,

d₁ = v₁t

d₁ = 18.2t

Initial speed of the second train = v₂ = 0 m/s (At rest)

Acceleration of the second train = a₂ = 0.99 m/s²

Distance traveled by the second train before the collision = d₂

d₂ = v₂t + a₂t²/2

d₂ = (0)t + (0.99)t²/2

d₂ = 0.495t²

The total distance traveled by both the trains before the collision is equal to the initial distance between the trains.

d = d₁ + d₂

242 = 18.2t + 0.495t²

0.495t² + 18.2t - 242 = 0

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

$t = \frac{-(18.2) \pm \sqrt{(18.2)^{2} - 4(0.495)(-242)}}{2(0.495)}$

$t = \frac{-18.2 \pm 28.4675}{0.99}$

t = 10.3712 sec or -47.139 sec

Time cannot be negative.

t = 10.3712 sec

d₁ = 18.2t

d₁ = 18.2(10.3712)

d₁ = 188.76 m

Distance the first train moves before the trains collide = 188.76 m