The escalator that leads down into a subway station has a length of 28.5 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?
time<11sec
(28.5)/(v-1.6) <11
28.5<11V-11*1.6
v> (28.5+11*1.6)/11
CHECK THAT.
To find the speed at which the student must exceed in order to beat the record, we need to consider the relative motion between the student and the escalator.
Let's assume that the student's speed relative to the ground (s) is greater than the speed of the escalator (e). Since the student is running in the wrong direction on the escalator, we can subtract the speed of the escalator from the student's speed to get the effective speed of the student relative to the ground.
Therefore, the effective speed of the student relative to the ground will be:
Effective speed = s - e
Now, let's calculate the time taken by the student to travel the length of the escalator. We know that the length of the escalator (L) is 28.5 m and the speed of the escalator (e) is 1.6 m/s.
Time taken to travel the length of the escalator = L / e
Now, we know that the local record time for this trick is 11 s. So, in order to beat this record, the time taken by the student to travel the length of the escalator should be less than 11 s.
We can set up the inequality:
Time taken to travel the length of the escalator < 11
L / (s - e) < 11
Now, let's substitute the given values:
28.5 / (s - 1.6) < 11
We can solve this inequality to find the range of values that the student's speed (s) must fall in to beat the record.