Question

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

Answer 1

Acceleration:
$a_x = \frac{d ^2x}{dt^2} \\ a_y = \frac{d ^2y}{dt^2} \\ a_x = 8 \\ a_y=0$

Since $a_x$ and $a_y$ are independent of time , at any time $a_x$ = 8 and $a_y$ = 0

Total acceleration is $\sqrt{8^{2} + 0^{2} } = 8 m/s^2$

Answer 2

$x = 7t + 4t^2\\ \\a_x= \frac{d^2x}{dt^2}= \frac{d}{dt^2}(7t+4t^2)=8m/s^2 \\ \\y = 5t \\ \\a_y= \frac{d^2y}{dt^2}= \frac{d}{dt^2}(5t)=0\\ \\at\ 5second,\\ a_x=8\ m/s\\a_y=0\\a= \sqrt{a_x^2+a_y^2}= \sqrt{8^2+0}=8\ m/s^2 \\ \\Acceleration\ of\ particle\ is\ 8m/s^2\ at\ the\ end\ of\ 5\ second.$