Question

The x and y coordinates of a particle at any time $t$ are given by $x=2t+4t^2$ and $y=5t$, where x and y are in meter and $t$ in second. The acceleration of the particle at $t=5s$ is:

• A. $40m{s}^{-2}$
• B. $20m{s}^{-2}$
• C. $8m{s}^{-2}$
• D. Zero

Answer 1

The correct option is D Zero

$x=2t+4t^2$

d.e.r. to $t$

$\frac{dx}{dt}=2+8t$

again d.e.r. to $t$

$\frac{d^{2}x}{d{t}^2}=8$

$y=5t$

$\frac{dy}{dt}=5$

$\frac{d^{2}y}{d{t}^2}=0$

$\because$ acceleration $=\frac{d^{2}y}{d{x}^2}=\frac{d^{2}y/d{t}^2}{d^{2}x/d{t}^2}=\frac{0}{8}$

acceleration $=0$

at $t=5$ acceleration $=0$ Answer

thus $D$ is the right answer.