Acar is moving in the positive direction along a straight highway and accelerates at a constant rate while going from point a to point b. if the acceleration is positive, increasing the speed of the car, when does the position where the average speed equals the instantaneous speed occur during the time interval from a to b? assume the time interval is t.

The time where the avergae speed equals the instaneous speed is T/2

Explanation:

The velocity of the car is:

v(t) = v0 + at

Where v0 is the initial speed and a is the constant acceleration.

Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:

v_ave = v0+a(T/2)

We can esaily note that when t=T/2

v(T/2)=v_ave

Now we want to know where the car should be, the osition of the car is:

Where x_A is the position of point A. Therefore, the car will be at:

x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.