Question

A particle moves along the parabolic path $y={ax}^2$ in such a way that the $y-$ component of the velocity remains constant, say $c$. The $x$ and $y$ coordinates are in meters. Then acceleration of the particle at $x=1m$ is:

• A. $ac\hat{k}$
• B. $2{ac}^2\hat{j}$
• C. $-\frac{c^2}{4a^2}\hat{i}$
• D. $-\frac{c}{2a}\hat{i}$

Answer 1

Answer: (B)

$2{ac}^2\hat{j}$

$y=a{x}^2$

Velocity is given by,

$\frac{dy}{dt}=2ax\frac{dx}{dt}$

As x component of the velocity remains constant i.e. $\frac{dx}{dt}=c$

$v_y=\left(2ac\right)x$

Acceleration is given by,

$\frac{d{v}_y}{dt}=2ac\frac{dx}{dt}$

$a_y=2a{c}^2$

Acceleration has only y component, so acceleration of the particle is in y-direction i.e.

$\overrightarrow{a}=2a{c}^2\overrightarrow{j}$