Question

A particle projected from origin to move in X-Y plane, with a velocity v=3i+6xj,where I,j are the unit vector along X and Y axis.find the equation of path following by the particle is .......

Answer 1

Answer on Question 48461, Physics, Mechanics | Kinematics | Dynamics

Question:

A particle projected from origin to move in X-Y plane, with a velocity $v = 3i + 6xj$, where $i, j$ are the unit vector along X and Y axis. Find the equation of path following by the particle is...

Solution:

The path following by the particle will be :

$$s = \sqrt {x ^ {2} + y ^ {2}},$$

where $x$ and $y$ are projections of the particle displacement on axes $x$ and $y$ respectively. To find the displacement of particle along axis $x$ and $y$ we use the second equation of motion:

$$s = v \cdot t + \frac {1}{2} \cdot a \cdot t ^ {2},$$

where $s$ is the displacement, $v$ is the initial velocity, $t$ is time, $a$ is acceleration. So, the displacement along axis $x$ would be:

$$x = 3 \cdot t.$$

The component of velocity along the $y$ axis would be: $v = 6j \cdot (3t) = 18 \cdot t$

Assuming that at $t = 0$ initial velocity $v = 0$ we obtain for displacement along axis $y$ :

$$y = \frac {1}{2} \cdot a \cdot t ^ {2} = \frac {1}{2} \cdot 18 \cdot t ^ {2} = 9 \cdot t ^ {2}.$$

Therefore, the equation of path following by the particle would be:

$$s = \sqrt {x ^ {2} + y ^ {2}} = \sqrt {\left(3 \cdot t\right) ^ {2} + \left(9 \cdot t ^ {2}\right) ^ {2}} = \sqrt {9 \cdot t ^ {2} \cdot \left(1 + 9 \cdot t ^ {2}\right)} = 3 \cdot t \cdot \sqrt {\left(1 + 9 \cdot t ^ {2}\right)}$$

Answer:

$$s = 3 \cdot t \cdot \sqrt {\left(1 + 9 \cdot t ^ {2}\right)}$$

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