Question

A particle projected from origin moves in x-y plane with a velocity v= 3i + 6xj, where i and j are the unit vectors along x and y axis. Find the equation of path followed by the particle:-

a) y=x2
b) y=1/x2
c) y=2x2
d) y=1/x

Answer 1

V=3i+6xj
V=dx/dt i + dy/dt j
dx/dt =3.........(1)
dy/dt=6x........(2)
Take eq.(1)....
dx=3.dt
Integrate both sides,we get-
x=3t .......(3)
Take eq.(2).....
dy=6x.dt
Put value of x from eq.(3)--
dy=6(3t).dt
dy=18t.dt
Integrate both sides,we get-
y=9t^2
Value of t from eq.(3) ..
t=x/3
y=9(x/3)^2
y=x^2

Answer 2

$\underline{\large\bf{\mathfrak{Hello!}}}$

To solve this question you need to know the concept of projectile and how we find out the angle formed by a projectile.
We should also know how to do integration.

$\boxed{y = {x}^{2}}$

$\bf{\mathfrak{Hope \: this \: helps...:)}}$