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

Grade:12th pass

## 1 Answers

Arun
25763 Points
3 years ago
You want the equation of the trajectory of the particle, only for its motion along the y axis.
Here, the velocity along the y axis is dependent on x.
So, we first find x for any given time t.
Using the second equation of motion,
s = ut + 1/2(a * t^2)
s = displacement, u = initial velocity, t = time, a = acceleration.
So, taking the displacement along x axis to be x,
x = 3t + 1/2(0 * t^2)
So, x=3t
Now, you get the component of velocity along the y axis in terms of t in which
v (along y) = 6(3t) j = 18t j
The next part involves calculus.
To find the acceleration, differentiate v with respect to t and you get
a = 18 j
So, you can find the displacement along the y axis ( take it as y) using the second equation again
y = ut + 1/2(a * t^2)
At t = 0 the initial velocity u will be 0
Therefore, y = 1/2 * a * t^2
And as a = 18
y = 9 * t^2

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