Question

A radius vector of point A relative to the origin varies with time t as $\vec r = at\hat i - b{t^2}\hat j$ where a and b are constants. Find the equation of the point's trajectory.
A. $y = \dfrac{{ - 4b}}{{{a^2}}}{x^2}$
B. $y = \dfrac{{ - 2b}}{{{a^2}}}{x^2}$
C. $y = \dfrac{{ - b}}{{{a^2}}}{x^2}$
D. $y = \dfrac{{ - b}}{{2{a^2}}}{x^2}$

Answer 1

Compare the given equation of point to the original equation of point lying on the x-y plane. Determine the x-coordinate and substitute the value of t from x-coordinate into the y-coordinate.

Complete step by step answer:
We have given the position of point A with respect to the origin with respect to time t as below,
$\vec r = at\hat i - b{t^2}\hat j$
Here, $a$ and $b$ are constants and t represents time.
We know that, the position of a certain point on x-y plane is represented as,
$\vec r = x\hat i + y\hat j$
Here, x is the x-coordinate of the point and y is the y-coordinate of the point r, $\hat i$ and $\hat j$ are the unit vectors along x and y axis respectively.
We can compare the above two equations to get the values of x and y as follows,
$x = at$ …… (1)
And,
$y = - b{t^2}$ …… (2)
From equation (1), we can write,
$t = \dfrac{x}{a}$
Substitute $t = \dfrac{x}{a}$ in equation (2).
$y = - b{\left( {\dfrac{x}{a}} \right)^2}$
$y = - \dfrac{b}{{{a^2}}}{x^2}$
This is the equation of trajectory.

From the above equation, we can say that the equation represents the shape of the parabola about the negative y-axis and $a$ and b are the constants.

So, the correct answer is option (C).

Note:
The position of a point on a x-y plane is represented as $\vec r = x\hat i + y\hat j$. If the x-coordinate is given negative, it should be taken as negative to substitute it in the y-coordinate and not the magnitude of that value.