Where Will the Ball Hit the Floor of the Railcart?

[supercali]

[SOLVED] a question

a railcart moves along a railway at constant velocity v_0 with a ball attached to its ceiling. at t=0 the ball is drooped from the ceiling and at the same time the railcart breaks are pushed creating deceleration such that its velocity is:
v=v_0e^{ -pt}\hat{x} .
h is the distance between the ceiling and the floor.
in which horizontal distance will the ball hit the floor of the railcart. given v_0=70m/sec
h=180m p=0.0185 1/sec g=10m/sec^2
if you can help me understand this and solve it
thanks

 

[Dick]

You will have to integrate the x velocity dt to get the horizontal displacement as a function of time. The vertical displacement is the usual (1/2)*g*t^2. Use that to find t when it hits the floor and put it into the x displacement.

 

[supercali]

clarifications

ok i understood
but when i integrate i have an x_0 is it zero or because the railcart is decelerating i need to find it.
and another question regarding the use of the well known equation x=v_0t+ \frac{at^2}{2} since there is an acceleration shoulnt there be a different accelaration according to the galilean transformation in accelerated systems

 

[supercali]

ok i did as you told me and got after integrating \frac{v_0e^{-pt}}{-p} and the answer wasnt right. i got -3386.25m fot the time 6sec
i think it has to do with the fact that i didnt calculate the x_0 how do i do that

 

[Dick]

ok i did as you told me and got after integrating \frac{v_0e^{-pt}}{-p} and the answer wasnt right. i got -3386.25m fot the time 6sec
i think it has to do with the fact that i didnt calculate the x_0 how do i do that

Your problem is that you can't just plug t=6 into that. You have to put t=6 and t=0 and then subtract them. It's an indefinite integral.

 

[Dick]

ok i understood
but when i integrate i have an x_0 is it zero or because the railcart is decelerating i need to find it.
and another question regarding the use of the well known equation x=v_0t+ \frac{at^2}{2} since there is an acceleration shoulnt there be a different accelaration according to the galilean transformation in accelerated systems

There is no 'x0' to find. The initial value of x doesn't matter. This is all about differences. And there is no need to change frames. You are doing fine solving the problem is this one.

 

[supercali]

**** i don't believe it i made that stupid mistake of not putting a zero in the integral

dude thak you very much
your the best
keep on helping it is the nicest thing ever