Particle 1 is moving on the x-axis with an acceleration of 5.55 m/s2 in the positive x-direction. Particle 2 is moving on the y-axis with an acceleration of 7.45 m/s2 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 3 s.
Answer in units of m/s
To find the speed of particle 1 with respect to particle 2 at 3 seconds, we need to calculate the velocity of each particle at that time and then find the relative velocity between them.
Particle 1:
Given that particle 1 has an initial velocity of 0 m/s and an acceleration of 5.55 m/s^2 in the positive x-direction, we can use the kinematic equation for linear motion:
v1 = u1 + a1 * t
where v1 is the final velocity of particle 1, u1 is the initial velocity (0 m/s), a1 is the acceleration (5.55 m/s^2), and t is the time (3 s).
Plugging in the values, we have:
v1 = 0 + 5.55 * 3
v1 = 16.65 m/s
Particle 2:
Similarly, particle 2 has an initial velocity of 0 m/s and an acceleration of 7.45 m/s^2 in the negative y-direction. Again using the above kinematic equation, we have:
v2 = u2 + a2 * t
where v2 is the final velocity of particle 2, u2 is the initial velocity (0 m/s), a2 is the acceleration (-7.45 m/s^2), and t is the time (3 s).
Substituting the values, we get:
v2 = 0 + (-7.45) * 3
v2 = -22.35 m/s
Relative velocity:
The relative velocity between particle 1 and particle 2 is the vector difference of their velocities. In this case, since particle 1 is moving only along the x-axis and particle 2 is moving only along the y-axis, their velocities are perpendicular to each other. Therefore, the relative velocity between them is the magnitude of their vector difference.
v_rel = sqrt(v1^2 + v2^2)
Substituting the values, we have:
v_rel = sqrt((16.65)^2 + (-22.35)^2)
v_rel ≈ 28.028 m/s
Therefore, the speed of particle 1 with respect to particle 2 at 3 seconds is approximately 28.028 m/s.