Question

The $x$ and $y$ co-ordinates of a particle at any time $t$ are given by $x=7t+4t^2$ and $y=5t$ where $x$ and $y$ are in m and $t$ in s. The acceleration of the particle at $5s$ is:-

• A. zero
• B. $8m/s^2$
• C. $20m/s^2$
• D. $40m/s^2$

Answer 1

The correct option is B $8m/s^2$

$x=7t+4t^{2}andy=5t$

$v_x=\frac{dx}{dt}=8t+7;v_y=\frac{dy}{dt}=5$
$a_x=\frac{d{v}_x}{dt}=8;a_y=\frac{d{v}_y}{dt}=0$

Now $a=\sqrt{a_x^2+a_y^2}=\sqrt{(8{)}^2+(0{)}^2}=8m/s^2$