escalator that leads down into a subway station has a length of 27.0 m and a speed of 1.7 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 the record?
If the student's speed relative to the escalator is s, then his speed relative to the ground is s-1.7. We want to cover the 27m length of the escalator in less than 11 seconds, so since time = distance/speed,
27/(s-1.7) < 11
s > 4.15 m/s
To determine the speed that the student must exceed in order to beat the record, we need to consider the speed of the escalator and the speed of the student.
Given information:
Length of the escalator (distance): 27.0 m
Speed of the escalator: 1.7 m/s
Record time: 11 s
Let's assume that the speed of the student relative to the ground is s, and the speed of the student relative to the escalator is v.
When the student is running in the wrong direction on the escalator, the speeds add up. Therefore, the total velocity (v_total) of the student relative to the ground is given by:
v_total = v + 1.7 m/s
The time taken by the student to cover the distance of the escalator is equal to the record time, which is 11 s. This can be calculated using the equation:
time = distance / velocity
11 s = 27.0 m / v_total
Substituting the value of v_total, we have:
11 s = 27.0 m / (v + 1.7 m/s)
To solve for v, we can cross-multiply and simplify:
11s(v + 1.7 m/s) = 27.0 m
11v + 18.7 m/s = 27.0 m
11v = 8.3 m
v ≈ 0.755 m/s
Therefore, the student must exceed a speed of approximately 0.755 m/s relative to the escalator in order to beat the record.
To solve this problem, we need to consider the relative velocity of the student with respect to the escalator. Let's break down the problem step by step:
1. The length of the escalator is 27.0 m, and it has a speed of 1.7 m/s relative to the ground. This means that in 1 second, the escalator moves 1.7 meters.
2. The student is running in the wrong direction on the escalator. To beat the record, the student needs to cover the length of the escalator faster than the escalator moves.
3. Let's assume that the student's speed relative to the escalator is 'v' m/s. In this case, the speed of the student with respect to the ground will be the sum of the speeds of the student and the escalator (since they are moving in the opposite direction). So the speed of the student with respect to the ground will be (1.7 + v) m/s.
4. To find the time taken by the student to cover the length of the escalator, we divide the length of the escalator by the speed of the student with respect to the ground. So the time taken by the student will be 27.0 m / (1.7 + v) m/s.
5. The record time for this trick is 11 seconds. If the student's time is less than 11 seconds, it means the student has beaten the record. So we need to find the speed 'v' that will make the time less than 11 seconds.
Let's set up an equation to solve for 'v':
27.0 m / (1.7 + v) m/s < 11 s
Now, we can solve this inequality to find the maximum value of 'v' that satisfies the condition.
27.0 / (1.7 + v) < 11
Solving this inequality will give us the speed 'v' that the student needs to exceed in order to beat the record.