Particles moving along the axes.

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Particle 1 accelerates at 5.45 m/s² along the x-axis, while Particle 2 accelerates at 7.56 m/s² in the negative y-direction, both starting from rest at the origin. After 6.15 seconds, the velocity of Particle 1 is calculated to be 33.5175 m/s, and Particle 2's velocity is -46.494 m/s. To find the speed of Particle 1 relative to Particle 2, the velocities are treated as perpendicular vectors. The resultant speed is determined using the Pythagorean theorem, yielding a speed of (v_1² + v_2²)^(1/2). The discussion clarifies the concept of relative velocity in a two-dimensional context.
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Particle 1 is moving on the x-axis with an acceleration of 5.45 m/s^2 in the positive x-direction. Particle 2 is moving on the y-axis with an acceleration of 7.56 m/s^2 in the negative y-direction. Both particles were at rest at the origin at t = 0 s. Find the speed of particle 1 with respect to particle 2 at 6.15 s. Answer in units of m/s.

I'm not sure what the question is asking for. The whole "with respect to" slightly confuses me. In the x-direction, the velocity after 6.15 is 33.5175. In the negative y-direction, it is 46.494. ...am I just suppose to draw the hypotenuse to the triangle formed and find that length?
 
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Yes. The velocity of #1 with respect to #2 is \vec{v}_1-\vec{v}_2. Since \vec{v}_1 is perpendicular to \vec{v}_2, the speed (which is the magnitude of \vec{v}_1-\vec{v}_2) is (v_1^2+v_2^2)^{\scriptstyle{1/2}}.
 
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