Here's the question you clicked on:
arcticf0x
A quick one, what will be the velocity-time graph of a projectile fired up with 20 m/s? Also if you can explain me how to feel the nature of motion from these v-t graphs, some of them have driven me nuts lately.
|dw:1330634460339:dw|That V=20 in the graph since the derivative of velocity is acceleration, which is a constant here (ignoring air resistance) Since the acceleration is equal to -g, that will be the slope of the velocity graph. I let g=10 in the drawing and used the fact that the velocity at the top is zero to find the time at the top of the motion t'\[v-gt'=0\to t'=\frac vg=2\]the total time will be double that The displacement will be given by the integral vdt, which is the area under the graph of v
*since the derivative of velocity is acceleration, which is a constant here (ignoring air resistance) the graph will be a straight line...
Ok, i got the straight line part, but am still confused about the initial point. For example, if a body at rest is dropped from top of a tower and comes to rest on striking the ground, what will be the v-t graph here then?
a body that starts a rest is what point on a v-t graph?|dw:1330791615547:dw|from the kinematic for free-fall we have\[h=\frac12gt^2\to t=\sqrt{\frac{2h}g}\]is the time is hits the ground the velocity when it hits the ground is\[v=gt=\sqrt{2gt}\]
|dw:1330793802169:dw|the slope of the line for\[0\le t\le \sqrt{\frac{2h}g}\]is -g Now the impact force of the ground on the object decelerates it. We can find that acceleration with\[F=ma\to a={F\over m}\]so the slope of the line after impact with the ground would be F/m
Aha, the eariler one had some initial velocity hence didnt start from the origin, but this one has u=0 hence starts from origin! Ok i got the slopes and the time thing! Thanks a lot TurningTest! xD