posted by Jenny .
For a question in physics... I was told to draw a graph of vertical displacement, y, versus time, t, in 2sec increments from 0 to 1.8. What was given to me was:
a_y_ = -10m/s^2
So I used the formula: y=vo+1/2at^2
I drew the graph, and got a parabala.
Now, my question is, how do I find instantaneous vertical velocity at 6s using the graph?
Do I use the linear question for horizontal motion at all in this problem? x=vo(t)
Any help would be great. THanks
To use the graph only, you could either (1) measure the tangent to the y(t) curve at = 6 s, or (with less accuracy):
(2) divide the difference in y altitudes at t = 7 s and t = 5s by the interval, 2 s.
You could also get the exact answer by differentiating the y(t) function itself, to get Vy (t), but they don't seem to want you to do that.
You will not need the horizontal equation of motion to do this problem.
Alright, thanks. I drew a tangent line and found the slope at t=6, and got 3.75 m/s as my instantaneous velocity.
Just so I know... when do I have to take into consideration the linear equation governing horizonatal motion?