1) 234 s, 6.18 km
The position of the car at time t is described as
[tex]x_c(t) = v_c t[/tex]
where
[tex]v_c = 95 km/h[/tex] is the velocity of the car
The position of the head of the train instead is given by
[tex]x_t(t) = d + v_t t[/tex]
where
d = 1.30 km is the initial distance between the car and the head of the train
[tex]v_t = 75 km/h[/tex] is the velocity of the train
The car overtakes the train when
[tex]x_c =x_t[/tex]
Substituting and solving for t,
[tex]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[/tex]
And the distance travelled by the car is
[tex]x_c = v_c t = (95 km/h)(0.065 h)=6.18 km[/tex]
2) 27.5 s, 0.72 km
In this case, the train is travelling in opposite direction, so we can write
[tex]v_t = -75 km/h[/tex]
Again, we can use the same equation as before
[tex]x_c =x_t[/tex]
And solving for t, we find
[tex]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[/tex]
And the distance travelled by the car is
[tex]x_c = v_c t = (95 km/h)(0.0076 h)=0.72 km[/tex]
