Question

An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track

Answer 1

1) 234 s, 6.18 km

The position of the car at time t is described as

$x_c(t) = v_c t$

where

$v_c = 95 km/h$ is the velocity of the car

The position of the head of the train instead is given by

$x_t(t) = d + v_t t$

where

d = 1.30 km is the initial distance between the car and the head of the train

$v_t = 75 km/h$ is the velocity of the train

The car overtakes the train when

$x_c =x_t$

Substituting and solving for t,

$v_c t = d + v_t t\\t(v_c-v_t)=d\\t=\frac{d}{v_c-v_t}=\frac{1.30}{95-75}=0.065 h = 234 s$

And the distance travelled by the car is

$x_c = v_c t = (95 km/h)(0.065 h)=6.18 km$

2) 27.5 s, 0.72 km

In this case, the train is travelling in opposite direction, so we can write

$v_t = -75 km/h$

Again, we can use the same equation as before

$x_c =x_t$

And solving for t, we find

$v_c t = d + v_t t\\t(v_c-v_t)=d\\t=\frac{d}{v_c-v_t}=\frac{1.30}{95+75}=0.0076 h = 27.5 s$

And the distance travelled by the car is

$x_c = v_c t = (95 km/h)(0.0076 h)=0.72 km$