The position of an objectmoving along x-axis is given by x = a + bt2where a = 8.5 m, b = 2.5 m s–2 and t ismeasured in seconds. What is its velocity att = 0 s and t = 2.0 s. What is the averagevelocity between t = 2.0 s and t = 4.0 s ?

The velocity of an object is given by the derivative of its position with respect to time.

Given the position function x = a + bt^2, the velocity function v(t) is the derivative of x with respect to time t.

v(t) = dx/dt = d(a + bt^2)/dt = 0 + 2bt = 2bt

1. To find the velocity at t = 0 s, we substitute t = 0 into the velocity function:

v(0) = 2b*0 = 0 m/s

So, the velocity of the object at t = 0 s is 0 m/s.

2. To find the velocity at t = 2.0 s, we substitute t = 2.0 into the velocity function:

v(2.0) = 2b2.0 = 4b = 42.5 = 10 m/s

So, the velocity of the object at t = 2.0 s is 10 m/s.

3. The average velocity between t = 2.0 s and t = 4.0 s is given by the change in position divided by the change in time.

The position at t = 2.0 s is x(2.0) = a + b*(2.0)^2 = 8.5 + 2.5*4 = 18 m

The position at t = 4.0 s is x(4.0) = a + b*(4.0)^2 = 8.5 + 2.5*16 = 48 m

So, the average velocity is (x(4.0) - x(2.0)) / (4.0 - 2.0) = (48 - 18) / 2 = 15 m/s

So, the average velocity between t = 2.0 s and t = 4.0 s is 15 m/s.