Let t be the time measured from when the velocity is 50 m/s
V = 50 + a t
If T is the time it takes to go from 50 to 80 m/s,
V(t) = 50 + 30 t/T
(The acceleration is a = 30/T)
Now integrate for X(t)
X(t) = 50 t + 15 t^2/T
X(T) = 50 T + 15 T = 65 T (point B location)
X(0) = 0 (point A location)
You want the time when X = X(T)/2
50 t + 15 t^2/T = 32.5 T
50 (t/T) + 15(t/T)^2 = 32.5
Solve for t/T
t/T = 0.556
V at that time is 50 + 0.556*30 = 66.7 m/s.. a bit more than the average speed
1)A car is moving with uniform accelaration. It moves along two point A and B and enters with velocity 50 m/s and 80 m/s respectively . Find the velocity when the car is moving in b/w the point A and B.
Here the body is not starting from rest. Now if it is in motion along staight line. If it enters at A at 50m/s and leaves B at 80 m/s that means body is accelerating. Now should I first find distance between A and B then calculate velocity or average velocity between A and B. which equation of motion can I use to solve this problem.
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