Question

Car 2 passes car 1. To a stationary observer, car 1 has a velocity of 28 m/s to the east and car 2 has a velocity of 33 m/s to the east. How much time will pass before car 2 is 14 m ahead of car 1?

Answer 1

`2.8\; {\rm s}`.

Step-by-step explanation:

The velocity of vehicle
`2` relative to vehicle
`1` is:

`\begin{aligned}& \text{velocity of 2 relative to 1} \\ =\; & (\text{velocity of 2 relative to ground}) \\ &- (\text{velocity of 1 relative to ground}) \\ =\; & 33\; {\rm m\cdot s^(-1) - 28\; {\rm m\cdot s^(-1)} \\ =\; & 5\; {\rm m\cdot s^(-1)} && (\text{to the east})\end{aligned}`.

The displacement of vehicle
`2` relative to vehicle
`1` is currently
`0\; {\rm m}`. The rate at which this displacement increases is equal to the velocity of vehicle
`2\!` relative to vehicle
`1\!`, which is
`5\; {\rm m\cdot s^(-1)}`.

Thus, it would take
`(14 \; {\rm m}) / (5\; {\rm m\cdot s^(-1)}) = 2.8\; {\rm s}` for this displacement to reach
`14\; {\rm m}`.