The x and y coordinates of a particle at any time t are given by `x=7t+4t^2` and `y=5t`, where x is in meter and t is in 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

B

Explanation

Given : ` x = 7t + 4t^(2) , y = 5t`
`:. v_x=(dx)/(dt)=7 + 8t ms^(-1) , v_y =(dy)/(dt) = 5 ms ^(-1)`
`a_x =(dv_x)/(dt)= 8 ms ^(-2) " "...(i)`
`a_y = (dv_y)/(dt)= 0" "...(ii)`
Equations (i) and (ii) show that acceleration does not depend on time i.e , the particle moves with constant acceleration a.
`a = sqrt((a_x)^3+(a_y)^2) = sqrt((8)^2 +(0)^2 ) = 8 ms ^(-2)`
Thus the acceleration of the particle at ` t = 5 s " is " 8 ms ^(-2)`