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

Hint: 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.