Question

A particle moves along the x axis according to the equation x = 2.08 + 3.06t − 1.00t^2, where x is in meters and t is in seconds. (a) Find the position of the particle at t = 3.30 s. (b) Find its velocity at t = 3.30 s. (c) Find its acceleration at t = 3.30 s.

Answer 1

Step-by-step explanation:

The position of a particle along x - axis is given by :

`x=2.08+3.06t-1t^2`

(a) Position at t = 3.3 s

`x=2.08+3.06(3.3)-1(3.3)^2`

x = 1.288 m

(b) Velocity at t = 3.3 s

`v=(dx)/(dt)`

`v=(d(2.08+3.06t-1t^2))/(dt)`

`v=3.06-2t`

at t = 3.3 s

`v=3.06-2(3.3)`

v = -3.54 m/s

(c) Acceleration,

`a=(dv)/(dt)`

`a=(d(3.06-2t))/(dt)`

`a=-2\ m/s^2`

Hence, this is the required solution.