Question

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.

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

For the question asked above this was the solution given but as I am not introduced to the integration concept is there any other method of solving this problem with the concepts I know that a class 9 students know.

Answer 1

The equation
X(t) = 50 t + 15 t^2/T
does not have to be derived with calculus.
It is just the familiar equation
X = Vo*t + (a/2)*t^2
In your case, Vo = 50 and a = (V2 - V1)/T

The rest of the derivation can proceed without calculus.