Topic:
Relative Velocity
Question:
The escalator that leads down into a subway station has a length of 30.0 m and a speed of 1.8 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?
(His speed-1.8m/s)*11sec=30m surely you can solve for his speed.
4.5
To solve this problem, we need to understand relative velocity and how it relates to the motion of the student and the escalator.
Relative velocity is the velocity of an object in a particular reference frame relative to another object or reference frame. In this case, we will consider the reference frame of the escalator.
Let's denote the speed of the student relative to the escalator as v, and the speed of the escalator relative to the ground as v_esc.
Now, let's break down the motion of the student and the escalator:
1. The speed of the student relative to the ground is the sum of their velocity relative to the escalator and the velocity of the escalator relative to the ground:
v_student = v + v_esc
2. The time it takes for the student to run on the escalator is t = 11 s.
3. The length of the escalator is 30.0 m.
Now, to beat the record, the student needs to cover the 30.0 m distance in less than 11 s. This can be achieved by running with a speed greater than the speed of the escalator.
Let's substitute the values given into the equation:
v_student = v + v_esc
v_student = v + 1.8 m/s
We want to find the speed of the student relative to the escalator that would allow him to beat the record. This means the distance covered in time t should be equal to or greater than 30.0 m:
d = v_student * t
30.0 m = (v + 1.8 m/s) * 11 s
Now, we can solve this equation to find the speed that the student must exceed in order to beat the record:
30.0 m = (v + 1.8 m/s) * 11 s
Divide both sides by 11 s:
2.73 m/s = v + 1.8 m/s
Subtract 1.8 m/s from both sides:
v = 2.73 m/s - 1.8 m/s
v = 0.93 m/s
Therefore, the student must exceed a speed of 0.93 m/s relative to the escalator to beat the record.