Question

object starts moving from rest on a horizontal ground such that position vector of the object at time T with respect to its starting point is given by r=4t^2i+(2t^-8t^4)j. The equation of trajectory of the particle at time t=1s will be:
a) 2y=x-x^2
b) y=x+x^2
c) y=2x-x^2
d) 2y=x-4x^2​

Answer 1

Answer:

The equation of trajectory of the particle is $2y = x - x^{2}$

Explanation:

The position vector of particle is given as

$\vec{r}= x i + y j = 4 t^{2} i + (2 t^{2} - 8 t^{4})j$

So

$x = 4 t^{2}$

$t^{2} = \frac{x}{4}$      ...1)

and $y= 2 t^{2} - 8 t^{4}$            ...2)

from equation 1 and equation 2

$y = 2\times \frac{x}{4} - 8\times \frac{x^{2}}{16}$

$y = \frac{x}{2} - \frac{x^{2}}{2}$

So  $2y = x - x^{2}$