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 t is seconds. The acceleration of particle at `t=5`s is

Options

A. zero
B. `8 ms^(-2)`
C. `20 ms^(-2)`
D. `40 ms^(-2)`

Answer

A

Explanation

`a=sqrt(a_(x)^(2)+a_(y)^(2))=[((d^(2)x)/(dt^(2)))^(2)+((d^(2)y)/(dt^(2)))^(2)]^(1//2)`
Here `(d^(2)y)/(dt^(2))=0`
Hence, `a=(d^(2)x)/(dt^(2))=8 ms^(-2)`