The x and y coordinates of the particle at any time are `x=5t-2t^(2) and y=10t` respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t=2s is:

Options

A. (a) `0`
B. (b) ` 5 m//s^2`
C. (c ) ` -4 m//s^2`
D. (d) - 8 m//s^2`

Answer

C

Explanation

`x=5 t- 2 t^2 ,`
velocity , ` v_x = (dx)/(dt) = 5- 4 t`
Acceleration, ` _x = (dv-x)/(dt) =5 - 4 t`
` y= 10 t`
Velocity, ` v_y = (dy)/(dt) = 10`
velocity, ` v_y = (dy)/(dt) = 10`
Acceleration, ` _y = (dv_y)/(dt) =0`
:. Acceleration of particle at ` t = 2 s`
` =sqrt ((-4)^2 + 0^2 ) =- 4 m//s^2`.