Question

The position of an object moving along x axis is given by x = a + bt^2 where
a = 8.5m, b = 2.5 m/s^2 and t is measured in seconds. What is its velocity at t=0 s and t = 2s. What is the average velocity between t = 2s and t = 4s ? ( NCERT Example 3.2 )

Answer 1

Answer:

V at t=0 is 0m\s, V at t=2 is 10m\s , and Vavg= 15m\s.

Explanation:

differentiating x,

so x'=2bt. (by differentiating x you will be getting velocity)

at t=0. V=0m\s

at t=2. V=10m\s

Now for Vavg= (X2 - X1)\(t2 - t1)

so Vavg=15m\s