Asked by Akashdeep

This question is based upon differential calculus. Velocity of the particle is given by the equation:
v = (2t^2+5)cm/s. Find:-

(i) The change in velocity of the particle during the time interval t1 = 2s and t2 = 4s.
(ii) Average acceleration during the same interval.
(iii) Instantenous acceleartion at t = 4s.

Please SOLVE it!!!
Answered by drwls
(i) Use the formula you havebeen given to compute v at t = 4 and t = 2 s. Compute the change in v:
v(4) - v(2) = 37 - 13 = ____ cm/s

(ii) Divide the result of (i) by 2 s.

(iii) Compute the formula for derivative of v(t).
It is a(t) = dv/dt = 4t.
That is the instantaneous acceleration. Evaluate it at t = 4s.

Fill in the blanks. You could use the exercise.
Answered by Akashdeep
Sir I want the answers so i could match my answers with yours
Answered by drwls
Then show your answers
Answered by Akashdeep
Sir i'll solve tommorrow and show you. I hope you get back to this question tommorow
Answered by drwls
I have already shown you how to do the three problems. Each requires one step. I will be happy to verify that you have followed directions
Answered by Akash
Sir my teacher solved it!!!
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