If a particle moving in straight line such that its position varies with time as x=5(t-2) + 6(t-2)^2 ,then intial acceleration is

12

Explanation:

x = 5 (t − 2) + 6 (t − 2)²

x = 5t − 10 + 6 (t² − 4t + 4)

x = 5t − 10 + 6t² − 24t + 24

x = 6t² − 19t + 14

Velocity is the derivative of position with respect to time:

v = dx/dt

v = 12t − 19

Acceleration is the derivative of velocity with respect to time:

a = dv/dt

a = 12

The particle's acceleration is constant at 12 (use appropriate units).

he kicked it 10 feet straight up into the air? i think this works.