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. 0
B. `5 m//s^(2)`
C. `-4m//s^(2)`
D. `-8 m//s^(2)`

Answer

C

Explanation

`x = 5t - 2t^(2) rArr v_(x) = (dx)/(dt) = 5-4t rArr a_(x) = 4`
`y = 10t rArr v_(y) = 10 rArr a_(y) ne 0`
`therefore veca =a_(x)hati + a_(y)hatj`
So, `veca = -4hatj`