91Ó°ÊÓ

Relative velocity is the speed of one object as observed from another object. In this exercise, we need to understand how fast Speedy Sue is approaching the van ahead of her. To find this, you subtract the velocity of the van from Sue's velocity.

Since Sue is traveling at \(30.0\, \mathrm{m/s}\) and the van at \(5.0\, \mathrm{m/s}\), the relative velocity is \(v_r = 30.0\, \mathrm{m/s} - 5.0\, \mathrm{m/s} = 25.0\, \mathrm{m/s}\).

This relative velocity is crucial because it tells us how quickly the gap between Sue and the van is closing. Understanding relative velocity helps in analyzing potential scenarios such as overtaking or, in this case, potential collisions.

Relative acceleration is how one object's acceleration compares to another's. It extends from relative velocity but considers changes in speed over time. Sue is decelerating because she is braking on a wet road, with an acceleration of \(-2.00\, \mathrm{m/s^2}\). Since the van maintains a constant speed, its acceleration is \(0\, \mathrm{m/s^2}\).

If we subtract the van's acceleration from Sue's, the relative acceleration \(a_r\) is \(-2.00 \mathrm{m/s^2} - 0\, \mathrm{m/s^2} = -2.00\, \mathrm{m/s^2}\).

Knowing relative acceleration is important when predicting motion changes. It tells us how quickly the difference in their velocities is shrinking. In this scenario, it shows how Sue is slowing down in relation to the stationary speed of the van.

To determine if a collision will happen, we need to first calculate how quickly Sue's relative speed allows her to cover the distance to the van. Given the initial gap of \(155\, \mathrm{m}\) between them and Sue's relative velocity of \(25.0\, \mathrm{m/s}\), the time to close this gap is \(t = \frac{155\, \mathrm{m}}{25.0\, \mathrm{m/s}} = 6.2\, \mathrm{s}\).

Next, we find out how long it takes for Sue to come to a complete stop using the formula \(t_{stop} = \frac{v_r}{-a_r}\). By calculation, \(t_{stop} = \frac{25.0\, \mathrm{m/s}}{2.00\, \mathrm{m/s^2}} = 12.5\, \mathrm{s}\). Since she can't stop in less time than it takes to reach the van (\(6.2\, \mathrm{s} < 12.5\, \mathrm{s}\)), a collision is inevitable. Knowing how to detect this gives us foresight into avoiding potential accidents.

Calculating the exact point of collision is the last step. We have determined the time it takes for Sue to catch up with the van (\(6.2\, \mathrm{s}\)). To find the collision point, we use the equation for distance covered under acceleration:

• \(d = v_{r}t + \frac{1}{2}a_{r}t^{2}\)

Substituting the values, we get:

\(d = 25.0\, \mathrm{m/s} \times 6.2\, \mathrm{s} + \frac{1}{2}\times (-2.00\, \mathrm{m/s^2})\times (6.2\, \mathrm{s})^{2} = 108\, \mathrm{m}\)

The collision happens \(108\, \mathrm{m}\) into the tunnel, \(6.2\, \mathrm{s}\) after Sue starts to brake.

This meticulous distance calculation is essential for understanding precisely where and when two objects will meet on a given trajectory.