The bus is 120 m ahead of the car, so the car covers this distance in
120 m = (30 m/s) <em>t</em>
<em>t</em> = (120 m) / (30 m/s)
<em>t</em> = 4 s
Now take the car's position at the moment the car overtakes the bus to be the origin.
Car's position at time <em>t</em> :
<em>x</em> = (30 m/s) <em>t</em>
Bus's position at time <em>t</em> :
<em>x</em> = (10 m/s) <em>t</em> + 1/2 (5 m/s²) <em>t</em>²
The bus overtakes the car when their positions are equal:
(30 m/s) <em>t</em> = (10 m/s) <em>t</em> + 1/2 (5 m/s²) <em>t</em>²
<em>t </em>/ (2 s) ((5 m/s) <em>t</em> - 40 m) = 0
<em>t</em> = 0 or <em>t</em> = (40 m) / (5 m/s) = 8 s
If the car travels a total distance of 450 m, then it does so after
450 m = (30 m/s) <em>t</em>
<em>t</em> = (450 m) / (30 m/s)
<em>t</em> = 15 s
The bus overtakes the car in the first 12 s of this journey, so both vehicles reach the same destination with 3 s between them.