Relative motion of an elevator question

[missrikku]

Okay, we're learning relative motion in class and I just want to see if I am uderstanding things correctly.

Okay there is a guy in an elevator and the elevator is on the ground. Also on the ground (outside of the elvator) is a ball. A ball is shot upward with Vo = 7 m/s. At the same time, the elevator moved up from the ground at a constant speec of Vc = 3 m/s. I have to find the maximum height (Ym) the ball reaches relative to a) the ground and b) the floor of the elevator? Then I have to find the rate at which the speedo f the ball changes relative to c) the ground and d) the elevator floor?

Well, this is what I am understanding:

a) relative to the ground, I can ignore the speed of the elevator so:
V^2 = Vo^2 + 2aYm
0 = 7 -19.6Ym
Ym = 0.357m

b) relative to the floor of the elevator cab, I am thinking that since the cab and the ball are both going up and the ball has a greater initial velocity:
V^2 = 0 = (Vo - Vc)^2 - 19.6Ym
0 = 4^2 - 19.6Ym
-16 = -19.6Ym
Ym = 0.816m

For c and d I am to find the rate at which the speed of the ball changes. Rate usually deals with time, right? And the change of speed over time is an acceleration, right?

Then can I do the following:

c) V(t) = Vot +0.5 at^2
0.357 = 7t -19.6t^2
t = 1.38s or t = 0.053s

Then can I take the initial Vo and divide by those times?

Vo/t1 = 7/1.38 = 5.07 m/s^2
Vo/t2 = 7/0.053 = 132 m/s^2

This sounds rather odd. Can someone point me the right direction? Thanks!

 

[HallsofIvy]

a) relative to the ground, I can ignore the speed of the elevator so:
V^2 = Vo^2 + 2aYm
0 = 7 -19.6Ym
Ym = 0.357m

Your formula is correct (It took me a moment to recognize it. It's from "conservation of energy) but since v0 is 7, you should have
0= 49- 19.6Ym, not "7".

b) relative to the floor of the elevator cab, I am thinking that since the cab and the ball are both going up and the ball has a greater initial velocity:
V^2 = 0 = (Vo - Vc)^2 - 19.6Ym
0 = 4^2 - 19.6Ym
-16 = -19.6Ym
Ym = 0.816m

Yes, this time you remembered to square!

For c and d I am to find the rate at which the speed of the ball changes. Rate usually deals with time, right? And the change of speed over time is an acceleration, right?

Yes, the (c) asks for the acceleration relative to the ground which you have ALREADY used. It is -g, the acceleration due to gravity.

In (d), since the elevator has no acceleration, the acceleration of the ball, relative to the elevator, should be the same.
You can calculate it directly by using the velocity formulas:
for the ball, v(t)= 7- gt. For the elevator, v(t)= 4.
The velocity of the ball, relative to the elevator, is
v(t)= (7- gt)- 4= 3- gt. Now find the acceleration, either by differentiating: v'= -g, or, since the acceleration is a constant by calculating the average acceleration over any time:
when t= 0, v= 4. When t= 3, v= 4- 3g. The change in speed is
4-3g- 4= -3g and, since this happened in 3 sec., the average acceleration is -3g/3= -g (of course).

 

[M98Ranger]

Your understanding of relative motion and the equations involved seem to be correct. However, there are a few things to clarify in order to fully answer the questions.

Firstly, for part a), the equation you have used is correct but the value for acceleration (a) should be positive since the ball is moving upwards. So the correct equation would be V^2 = Vo^2 + 2aYm, where a = 9.8 m/s^2.

For part b), you are correct in considering the initial velocity of the ball to be 7 m/s and the relative velocity of the elevator to be 3 m/s. However, the equation you have used is not correct. The correct equation would be V^2 = (Vo + Vc)^2 - 2aYm, where a = 9.8 m/s^2. This is because the elevator is also moving upwards and will affect the motion of the ball.

For parts c) and d), you are correct in using the equations V(t) = volt + 0.5at^2 and Vo/t = a. However, you need to consider the relative velocity of the ball with respect to the ground and the elevator floor separately.

For part c), you can use the equation V(t) = volt + 0.5at^2 to find the time (t) when the ball reaches its maximum height relative to the ground. Then, you can use the equation Vo/t = a to find the acceleration (a) of the ball relative to the ground.

For part d), you can use the same equation V(t) = volt + 0.5at^2 to find the time (t) when the ball reaches its maximum height relative to the elevator floor. Then, you can use the equation Vo/t = a to find the acceleration (a) of the ball relative to the elevator floor.

In summary, your understanding of relative motion and the equations involved is correct. Just make sure to consider the relative velocities and accelerations separately for the ground and the elevator floor. Keep up the good work!

 

[M98Ranger]

Your understanding is mostly correct, but there are a few things to clarify. First, for part a, you are correct in ignoring the speed of the elevator as it does not affect the motion of the ball. However, your calculation for Ym is incorrect. The correct equation to use would be V^2 = Vo^2 + 2aYm, where V is the final velocity (0 in this case). So the correct calculation would be 0 = 49 - 19.6Ym, which gives Ym = 2.5m. This is the maximum height the ball reaches relative to the ground.

For part b, your calculation is correct, but you should note that the initial velocity for the ball relative to the elevator floor is 4 m/s, not 7 m/s. So the correct equation would be 0 = 16 - 19.6Ym, which gives Ym = 0.816m.

For parts c and d, you are correct in using the equation V(t) = volt + 0.5at^2 to find the time at which the ball reaches its maximum height. However, the values you calculated for t are incorrect. The correct time for part c would be t = 0.714s, and for part d it would be t = 0.053s. These values can then be used to calculate the rate of change of speed, which is the acceleration.

So for part c, the acceleration of the ball relative to the ground would be Vo/t1 = 7/0.714 = 9.8 m/s^2, which is the acceleration due to gravity. And for part d, the acceleration of the ball relative to the elevator floor would be Vo/t2 = 4/0.053 = 75.47 m/s^2, which is the sum of the acceleration due to gravity and the acceleration of the elevator.

I hope this helps clarify things for you. Keep up the good work!