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!!!
(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.
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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
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To solve the given problem, we need to use differential calculus. Let's break down each part of the question and find the answers step by step:
(i) The change in velocity of the particle during the time interval t1 = 2s and t2 = 4s.
The change in velocity is obtained by finding the difference between the velocities at t2 and t1.
Given that v = (2t^2 + 5) cm/s, we substitute t2 = 4 into the equation: v2 = (2(4^2) + 5) cm/s = 41 cm/s.
Next, substitute t1 = 2 into the equation: v1 = (2(2^2) + 5) cm/s = 17 cm/s.
Finally, the change in velocity is calculated as Δv = v2 - v1 = 41 cm/s - 17 cm/s = 24 cm/s.
Therefore, the change in velocity during the time interval t1 = 2s and t2 = 4s is 24 cm/s.
(ii) The average acceleration during the same interval.
To find the average acceleration, we use the formula a = Δv/Δt, where Δv is the change in velocity, and Δt is the change in time.
Using the values we just calculated, Δv = 24 cm/s, and Δt = t2 - t1 = 4s - 2s = 2s.
Now, substitute these values into the formula: a = Δv/Δt = 24 cm/s / 2s = 12 cm/s².
Therefore, the average acceleration during the time interval t1 = 2s and t2 = 4s is 12 cm/s².
(iii) The instantaneous acceleration at t = 4s.
To find the instantaneous acceleration at t = 4s, we differentiate the velocity function with respect to time (t).
Given v = 2t^2 + 5 cm/s, differentiate it with respect to t:
dv/dt = d/dt(2t^2 + 5)
= 4t.
Now, substitute t = 4 into the equation: a = 4t = 4(4) = 16 cm/s².
Therefore, the instantaneous acceleration at t = 4s is 16 cm/s².
In summary:
(i) The change in velocity of the particle during the time interval t1 = 2s and t2 = 4s: 24 cm/s.
(ii) The average acceleration during the same interval: 12 cm/s².
(iii) The instantaneous acceleration at t = 4s: 16 cm/s².