Question

let x(t)= t2/3 give the distance of a moving particle from its starting point as a function of time t. for what value of t is the instanteous velocity of the particle equal to its average velocity over the interval (0,8)

Answer 1

To find the value of t at which the instantaneous velocity of the particle is equal to its average velocity over the interval (0,8), we need to first determine the instantaneous velocity and the average velocity.

The instantaneous velocity of the particle is the derivative of the distance function with respect to time, given by:

v(t) = dx/dt = 2/3 * t^(-1/3)

The average velocity of the particle over the interval (0,8) is the total distance traveled divided by the total time elapsed:

average velocity = (distance traveled) / (time elapsed)

= x(8) / 8

We can find x(8) by plugging t = 8 into the distance function:

x(8) = (8^(2/3)) = 4

Therefore, the average velocity over the interval (0,8) is 4/8 = 1/2.

Now, we need to find the value of t for which the instantaneous velocity is equal to 1/2. Setting v(t) equal to 1/2 and solving for t, we get:

2/3 * t^(-1/3) = 1/2

t^(-1/3) = 3/4

Taking the cube of both sides, we get:

t = (4/3)^3 = 64/27

Therefore, the instantaneous velocity of the particle is equal to its average velocity over the interval (0,8) when t = 64/27.