Help verify this question about average acceleration?

AI Thread Summary
The average acceleration of a car that increases its velocity from 5.51 m/s to 8.09 m/s over 7.36 seconds is calculated using the formula avg. acceleration = (v2 - v1) / (t2 - t1), resulting in 0.350 m/s². There was initial confusion regarding the use of "0s" for the starting time, but it was clarified that this is acceptable as it represents the beginning of the time interval. The discussion emphasized understanding the change in velocity over the time interval as Δv/Δt. Participants confirmed the approach and calculations were correct. The final consensus reassured that the method used to determine average acceleration was appropriate.
iamjohnny56
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Homework Statement



A car traveling in a straight line has a velocity of 5.51 m/s at some instant. After 7.36 s, its velocity is 8.09 m/s.
What is its average acceleration in this time interval?
Answer in units of m/s^2.

Homework Equations



avg. acceleration = (v2 - v1) / (t2 - t1)

The Attempt at a Solution



(8.09 m/s) - (5.51 m/s) / (7.36 s - 0 s)
= .350 m/s^2

---------------

The only thing I'm not sure about is the "0s" used above: is that the right interval? I'm confused if I'm approaching this right, so could anyone help me please?

Thanks!
 
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Looks perfect to me. Think of it as a = Δv/Δt, where the Δt is the time interval during which the velocity changes.
 
iamjohnny56 said:

Homework Statement



A car traveling in a straight line has a velocity of 5.51 m/s at some instant. After 7.36 s, its velocity is 8.09 m/s.
What is its average acceleration in this time interval?
Answer in units of m/s^2.

Homework Equations



avg. acceleration = (v2 - v1) / (t2 - t1)

The Attempt at a Solution



(8.09 m/s) - (5.51 m/s) / (7.36 s - 0 s)
= .350 m/s^2

---------------

The only thing I'm not sure about is the "0s" used above: is that the right interval? I'm confused if I'm approaching this right, so could anyone help me please?

Thanks!

Yes, it's fine to call the start time t=0. You could also just call the denominator delta_t=7.36s.
 
Ok, thank you guys! It does make more sense to think of it as Δv/Δt, so thank you!
 
Oh, man. Doc beat me to it again. Sigh.
 
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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