The position of an object moving along x -axis is given by x = a +`bt^(2)` where a = 8.5 m , b= 2.5 `m//s^(2)` and t is measured in seconds. What is its velocity at t=0 and t = 2.0 s? What is the average velocity between t = 2.0 s and t = 4.0 s?

Explanation

Given x = a +`bt^(2)`
Now instantaneous velocity v = `(dx)/(dt)= `2bt
So, at t =0 , `(dx)/(dt)|_(t=0)=0,`
at t =2s , `(dx)/(dt)|_(t=2)` = 4b = 10 m/s
Now, at t = 2 s, `x_(1)= a+4b`
at t = 4 s , `x_(2)` = a+16 b
`:.` Average velocity = `(x_(2)-x_(1))/(t_(2)-t_(1)) = ((a+16b)-(a+4b))/(4-2)`
` (12b)/(2)= 15 m//s[because`b =2.5]