Stumped on physics question, please help!?

Question

The escalator that leads down into a subway station has a length of 30.0m and a speed of 1.8 m/s relative to the ground. A student is coning out of the station and running up it in the wrong direction. The local record time for this stunt is 11s. Relative to the escalator what speed must the student exceed in order to beat the record?

Please show work. thanks so much!

Answer 1

If the elevator was still, you would need to travel faster than 30 meters in 11 Seconds. 30/11=2.73 m/s to meet the record.

Now because the elevator is working against you at a rate of 1.8 m/s, you need to overcome this.

2.73 + 1.8 = 4.53 m/s will match the record. Faster will beat it.

Answer 2

If a student is running up the escalator in the opposite direction, this means his velocity (relative to the escalator) opposes the velocity of the escalator (relative to the ground). The equation for the magnitude of the relative velocities yields:
v(SG) = v(SE) ─ v(EG)
v(SG) = velocity of the student with respect to the ground
v(SE) = velocity of the student relative to the escalator
v(EG) = velocity of the escalator relative to the ground
v(SE) = v(SG) + v(EG)
We can calculate the student's velocity relative to the ground from the length of the escalator and his record time:
v(SG) = 30.0 m/11 s = 2.727 m/s and
v(SE) = 1.8 m/s + 2.7 m/s = 4.5 m/s

Answer 3

by using fact the shell is max top, it has an preliminary velocity of 0. Then to drop sixteen.40 8.... m, x = a million/2 gt^2, the place x is sixteen.40 8 and g is 9.8 m/sec2. this might desire to be some million.4 sec. the fee whilst it hits the floor is v= gt, the place g is 9.8 m/sec^2 and t = a million.4 sec.

Answer 4

distance = vel * time. 30 = vel * (11), vel = 2.727... would be absolute "speed".. relative to the escalator, vel > 2.727 + 1.8 = 4.5727 m/s