The x and y-coordinates of a particle at any time t are given by `x=7t+4t^(2)` and y = 5t, where x and y are in metre and t seconds. The acceleration of particle at t = 5 is

Options

A. zero
B. `8m//s^(2)`
C. `20m//s^(2)`
D. `40m//s^(2)`

Answer

B

Explanation

`x=7 t+4t^(2) v_(x)=(dx)/(dt)=7+4xx2t`
`a_(1)=(dv_(x))/(dt)=8 m//sec^(2)`
`y=5 t rArr v_(y)=(dy)/(dt)=5` and `a_(y)=0`
`therefore a=sqrt(a_(x)^(2)+a_(y)^(2))=sqrt((8)^(2)+0)=8 m//s` does not depend on time.