Question

A particle moves on the curve $y = \dfrac{{{x^2}}}{4}$ where $x = \dfrac{t}{2}$. x and y are measured in meter and t in second. At $t = 4s$, the velocity of particle is:
A) $\dfrac{1}{{\sqrt 2 }}$
B) $\sqrt 3 $
C) $\sqrt 2 $
D) $2\sqrt 2 $

Answer 1

Hint: According to the question we have to find the velocity of a particle at $t = 4s$. For this, we will substitute the value of $a$ in \[y\] co-ordinate. Then we will find the velocity component in \[x\] and \[y\] coordinates. By substituting the value of t in the velocity component we can get the velocity of the particle.

Formulas used:
${v_x} = \dfrac{{dx}}{{dt}}$
${v_y} = \dfrac{{dy}}{{dt}}$
${\text{V = }}\sqrt {v_x^2 + v_y^2} $.

Complete step by step answer:
As per the question, we have \[x\] and \[y\] coordinates as
$\;x = \dfrac{t}{2},y = \dfrac{{{x^2}}}{4}$.
The velocity component in \[x\]direction is
${v_x} = \dfrac{{dx}}{{dt}} = \dfrac{1}{2}{\text{m/s}}$
Let’s substitute the value of \[x\] coordinate in the value of y coordinate,
$y = \dfrac{{{t^2}}}{{16}}$
The velocity component in \[y\] direction is
${v_y} = \dfrac{{dy}}{{dt}} = \dfrac{{2t}}{{16}} = \dfrac{t}{8}{\text{m/s}}$
At $t = 4s$, ${v_y} = \dfrac{4}{8} = \dfrac{1}{2}{\text{m/s}}$
Then the net velocity can be given by the,
${\text{V = }}\sqrt {v_x^2 + v_y^2} = \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{2}} \right)}^2}} = \dfrac{1}{{\sqrt 2 }}m/s$

Additional information:
The by-product of a function representing the position of a particle on a line at a time is the instantaneous velocity. The by-product of the velocity, which is the second by-product of the position function, represents the acceleration of the particle at time t.The average velocity between two points on the path in the limit that the time between the two points approaches zero.

Note: A positive acceleration shows that the velocity is increasing with respect to time, and the negative acceleration implies that the velocity is decreasing with respect to time. If the velocity remains constant on an interval of time, then the acceleration will be zero in the interval.