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Position of a particle is given by. X= 3t-4(t)^2+t^3 X is distance in metres and t is time in secs Find the average velocity of the particle in the time interval from t=2sec to t=4sec. Sir/ma'am I understood the method in my book which said that find displacement by subtracting x(4)-x(2) and then dividing it by 4-2. But I feel it we should also get the answer when we differentiate x w.r.t t and substitute the values and divide by 2. (dx/dt(4)+dx/dt(2))/2. But I am not getting the same answer. Pls explain why I am wrong, Thanks Jai

Position of a particle is given by. X= 3t-4(t)^2+t^3
X is distance in metres and t is time in secs
Find the average velocity of the particle in the time interval from t=2sec to t=4sec.
Sir/ma'am I understood the method in my book which said that find displacement by subtracting x(4)-x(2) and then dividing it by 4-2.
But I feel it we should also get the answer when we differentiate x w.r.t t and substitute the values and divide by 2.
(dx/dt(4)+dx/dt(2))/2.
But I am not getting the same answer. Pls explain why I am wrong,
Thanks 
Jai

Grade:12

1 Answers

Apoorva Arora IIT Roorkee
askIITians Faculty 181 Points
9 years ago
Whenever you are supposed to find the average velocity, What you need to do is take account of the total distance mentioned and then divide it by the mentioned time.
However if you differentiate, you get the instantaneous velocity at that point and Calculating instantaneous velocity at the extreme points i.e. at 2 sec and then at 4 sec and then dividing it by two can never give you the average velocity along the whole path, it will only give you the mean velocity of the two extreme points.
I hope you got the point,

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