Determine acceleration from position vs time graph

AI Thread Summary
To determine acceleration from a position vs. time graph, it's essential to accurately estimate instantaneous velocities at key points, such as half the height and the climax. The slopes of these points are used to create a velocity vs. time graph. The slope of the velocity graph represents acceleration. If the calculations are consistently incorrect, it may be due to misinterpretation of the graph or errors in slope calculation. Clarifying specific questions or issues can lead to better guidance.
Momentum09
Messages
69
Reaction score
0
Hi,
I was given a position vs. time graph of a projectile. I estimated the instantaneous velocity at the first 1/2 height, climax, and the second 1/2 height. From these slopes I created a velocity vs. time graph and use the slope of this line for my acceleration, but I keep getting this problem wrong.

Thanks!
 
Physics news on Phys.org
OK,

Now if you want some help you will need to ask a question.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

Similar threads

Back
Top