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From the origin, a particle starts at t ...

From the origin, a particle starts at t = 0 s with a velocity `vecv=7.0hatim//s` and moves in the xy plane with a constant acceleration of `veca=(-9.0hati+3.0hatj)m//s^(2)`. At the time the particle reaches the maximum x coordinate, what is its (a) velocity and (b) position vector?

Text Solution

AI Generated Solution

To solve the problem step by step, we will analyze the motion of the particle in both the x and y directions separately. ### Given Data: - Initial velocity: \(\vec{v}_0 = 7.0 \hat{i} \, \text{m/s}\) - Acceleration: \(\vec{a} = -9.0 \hat{i} + 3.0 \hat{j} \, \text{m/s}^2\) ### Step 1: Determine the time at which the particle reaches the maximum x-coordinate. At the maximum x-coordinate, the final velocity in the x-direction (\(v_{fx}\)) will be 0. We can use the equation of motion: ...
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Knowledge Check

  • A particle starts from the origin at t= 0 s with a velocity of 10.0 hatj m//s and moves in the xy -plane with a constant acceleration of (8hati+2hatj)m//s^(-2) . Then y -coordinate of the particle in 2 sec is

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