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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

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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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3 years ago
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