- #1
rainstom07
- 16
- 0
I just need some clarification.
In a lab that i did, a metal ball goes down a slanted ramp, onto a horizontal ramp of d distance, and then goes airborne (off the table). The instant the ball enters the horizontal ramp a timer is started, and - when it goes airborne - the timer is stopped. Thus, i have t time.
When i calculate using the formula [tex]v = d/t[/tex], i am calculating instantaneous velocity of the ball the instant it goes airborne correct?
When i consider where the ball goes onto the of horizontal ramp to be the zero point, i can say that the ball has displaced d distance, right? And using the same logic with the timer, i can say that the change in time is t time, correct? Thus i have Δd and Δt.
Using the formula [tex]v_{avg} = \Delta d/ \Delta t[/tex], yields me the average velocity correct?
Thanks in advance.
In a lab that i did, a metal ball goes down a slanted ramp, onto a horizontal ramp of d distance, and then goes airborne (off the table). The instant the ball enters the horizontal ramp a timer is started, and - when it goes airborne - the timer is stopped. Thus, i have t time.
When i calculate using the formula [tex]v = d/t[/tex], i am calculating instantaneous velocity of the ball the instant it goes airborne correct?
When i consider where the ball goes onto the of horizontal ramp to be the zero point, i can say that the ball has displaced d distance, right? And using the same logic with the timer, i can say that the change in time is t time, correct? Thus i have Δd and Δt.
Using the formula [tex]v_{avg} = \Delta d/ \Delta t[/tex], yields me the average velocity correct?
Thanks in advance.