This question is based upon differential calculus. Velocity of the particle is given by the equation:

Question

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!!!

Answer 1

(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.

Answer 2

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Answer 3

Then show your answers

Answer 4

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Answer 5

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

Answer 6

Sir my teacher solved it!!!