The escalator that leads down into a subway station has a length of 29.5 m and a speed of 1.6 m/s relative to the ground. A student is coming out of the station by running in the wrong direction on this escalator. The local record time for this trick is 11 s. Relative to the escalator, what speed must the student exceed in order to beat this record?

student speed = s

student speed over ground = (s-1.6)
29.5 = (s-1.6)(11)
solve for s

1.67

To determine the speed the student must exceed relative to the escalator in order to beat the record, we need to consider the relative velocities involved.

Let's break down the problem step by step:

Step 1: Find the velocity of the student relative to the escalator.
Since the student is running in the opposite direction to the escalator, the velocity of the student relative to the escalator is the difference between their velocities. The escalator has a speed of 1.6 m/s, so the velocity of the student relative to the escalator is:

Velocity of student relative to escalator = 0 m/s (student's running speed) - 1.6 m/s (escalator's speed) = -1.6 m/s

The negative sign indicates that the student is running in the opposite direction to the escalator.

Step 2: Determine the distance the student needs to cover.
The length of the escalator is given as 29.5 m. However, the student needs to run down the escalator in the opposite direction, so the distance the student needs to cover is 2 times the length of the escalator:

Distance = 2 * 29.5 m = 59 m

Step 3: Calculate the time the student needs to exceed.
The local record time for this trick is given as 11 s.

Step 4: Use the formula to calculate the speed of the student relative to the escalator.
Speed = Distance / Time

Speed = 59 m / 11 s ≈ 5.36 m/s

Therefore, the student must exceed a speed of approximately 5.36 m/s relative to the escalator in order to beat the record.