Finding Instantaneous acceleration from a velocity-time graph

AI Thread Summary
To find instantaneous acceleration from a velocity-time graph at t2 = 31 s, the slope of the tangent line at that point must be calculated. The user attempted various values but did not achieve the correct answer, indicating confusion over the conversion to m/s². They correctly computed average accelerations for different intervals but struggled with the instantaneous value at the corner point. The user concluded that the instantaneous acceleration is not zero and recognized that the slope at the corner is defined. Ultimately, they resolved their confusion regarding the calculation.
MitsuShai
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Graph: http://s324.photobucket.com/albums/k327/ProtoGirlEXE/?action=view&current=Capture.jpg
What is the instantaneous acceleration at t2 = 31 s?

I know that the instantaneous acceleration is the slope of the tangent of that point and yes I do know I have to convert it to (m/s^2). But I am still not getting the right answers. Here are a list of answers I imputed and they are all wrong:
-.1
0
-2
-1
-1.9
1.9
-.278
-.52

on the first few, I forgot to convert. I used 58/31 the last time...

The question had parts to it, but I don't know if they are necessary to solve this problem...here it is anyways:
A) Compute the average acceleration during the time interval t = 0s to t = 10s. 1.7
B) Compute the average acceleration during the time interval t = 30s to t = 40s. -1.7
C) Compute the average acceleration during the time interval t = 10s to t = 30s. 0
What is the instantaneous acceleration at t1 = 29 s. 0
I already computed this and got them right.
 
Last edited:
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It looks like its either going to be 0 or its going to be at that corner, where the rate of change is not going to be defined.
 
Whitishcube said:
It looks like its either going to be 0 or its going to be at that corner, where the rate of change is not going to be defined.

I put zero and it was wrong and I don't think undefined is an appropriate answer because 30 is exactly at that corner and it has an acceleration, which I used to solve part B with.
 
Graph: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/Capture.jpg
 
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nevermind I got it now.
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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