Assume you are on a planet similar to Earth where the acceleration of gravity os

Question

Assume you are on a planet similar to Earth where the acceleration of gravity os approximately 10m/s^ and the positive directions for displacement, velocity, and acceleration are upward. At time=0 s, an elevator is at a displacement of x=0m with a velocity of v= 0m/s. A student whose normal weight is 400N stands on a scale reading as a function of time. What is the displacement of the elevator above the starting point at the end of the 4th 5 s interval (at 20 s) answer in units of m. Graph shows scale reading N on y axis from 0-800 in increments of 200 and time (s) on X axis from 0-20 in increments of 5 readings are starts at 400N at 0 s to 5 s, then from 5 s to 10s goes to 500 N, from 10s to 15 s goes to back down to 400 N, and from 15 s to 20 s it goes up to 600 N.

What is the displacement of the elevator
above the starting point at the end of the
fourth 5 s interval (at 20 s)?
Answer in units of m

Explanation / Answer

After five seconds the elevator is in the same position. Based on the acceleration and weight specified we can say that the student has m=G/g=40Kg The first five seconds there is no acceleration of the elevator, so it stands still... The next five seconds, the student feels 100N hevier so the elevator has some acceleration. So the acceleration of the student was a=F/m=100/40=2.5m/s^2 The final velocity is v(10)=a*(t-5)=12.5m/s The final displacement is: d(t)=0+0*(t-5)+(a*(t-5)^2)/2=31.25m The next five seconds there is no acceleration, so the elevator sises with constant velocity a=0 v(15)=V(10)=12.5m/s d(t)=d(10)+v*(t-10)=31.25+62.5=93,75m/… The last portion there is an extra 200 of force, so the acceleration of hte elevator is a=200/40=5m/s^2 the fianl displacement is: d(t)=d(15)+v(15)*(t-15)+(a*(t-15)^2)/2…

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