Question

A particle is moving in y-z plane starting from origin.
The velocity component of the parudle along y-axis and
z-axis is given v in y= 3 m/s and v, = (2 + 3t(s)) m/s
The path of the particle is​

Answer 1

Answer:

First find position vectors in y and z directions

$x = \int \: v.dt$

in y - direction

$y = \int \: 3dt = 3t + c \\ t = 0 \: \: \: \rightarrow \: y = 0 \\ c = 0 \\ \\ t = \frac{y}{3}$

in x - direction

$z = \int \: (2 + 3t)dt = 2t + \frac{3}{2} {t}^{2} + c \\ at \: \: t = 0 \: \: \rightarrow \: \: z = 0 \implies \: c = 0\\ \\2 z = 3 {t}^{2} + 4t$

Put t = y/3

$2z = 3 {( \frac{y}{3}) }^{2} + 4 \times \frac{y}{3} \\ \\ 6z = {y}^{2} + 4y \\ \\ z = \frac{y(y + 4)}{6}$

So path of particle is

$\huge{Parabolic}$