Question

A car starts moving from rest on a horizontal ground such that the position vector of car with respect to its starting point is given as $(\vec{r} = bt\hat{i} + ct^2\hat{j})$, where $a$ and $b$ are positive constants. The equation of trajectory of car $y = f(x)$ is

• A. $y = (\frac{c}{b})x^2$
• B. $y = (\frac{c}{b^2})x^2$
• C. $y = (\frac{c^2}{b})x$
• D. $y = (\frac{c}{b^2})x$

Answer 1

Answer: (B)

$y = (\frac{c}{b^2})x^2$

Answer 2

The position vector of the car is given by:

$\vec{r} = bt\hat{i} + ct^2\hat{j}$

From this, we can identify the x and y components of the position:

$x = bt$ $y = ct^2$

To find the equation of the trajectory $y = f(x)$, we need to eliminate the time variable $t$ from these two equations.

From $x = bt$, we can express $t$ in terms of $x$:

$t = \frac{x}{b}$

Now, substitute this expression for $t$ into $y = ct^2$:

$y = c \left(\frac{x}{b}\right)^2$ $y = c \frac{x^2}{b^2}$ $y = \left(\frac{c}{b^2}\right)x^2$

This is the equation of the trajectory of the car.