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.
To find the speed of particle 1 with respect to particle 2 at 3 s, we need to calculate the velocities of each particle at that time and then calculate the relative velocity between them.
We can start by finding the velocity of particle 1 at 3 s using the equation:
v1 = u1 + a1 * t
Where:
v1 = velocity of particle 1
u1 = initial velocity of particle 1 (which is 0 m/s since it was at rest at the origin)
a1 = acceleration of particle 1 (5.55 m/s^2)
t = time (3 s)
Plugging in the values:
v1 = 0 + 5.55 * 3
v1 = 16.65 m/s
Now, we can find the velocity of particle 2 at 3 s using a similar equation:
v2 = u2 + a2 * t
Where:
v2 = velocity of particle 2
u2 = initial velocity of particle 2 (which is 0 m/s since it was at rest at the origin)
a2 = acceleration of particle 2 (-7.45 m/s^2) (negative because it's in the negative y-direction)
t = time (3 s)
Plugging in the values:
v2 = 0 + (-7.45) * 3
v2 = -22.35 m/s
Now we can calculate the speed of particle 1 with respect to particle 2:
speed = √((v1 - v2)^2)
speed = √((16.65 - (-22.35))^2)
speed = √(39^2)
speed = √(1521)
speed = 39 m/s
Therefore, the speed of particle 1 with respect to particle 2 at 3 s is 39 m/s.
To find the speed of particle 1 with respect to particle 2 at 3 seconds, we need to first find the velocities of each particle at that time.
For particle 1, we know it has an acceleration of 5.55 m/s^2 in the positive x-direction. Since it starts from rest at the origin, we can use the equation of motion:
V₁ = V₀ + a₁t,
where V₁ is the velocity of particle 1 at time t, V₀ is the initial velocity (which is zero in this case), a₁ is the acceleration of particle 1, and t is the time (3 seconds in this case).
Using the values given, we substitute them into the equation:
V₁ = 0 + (5.55 m/s²)(3 s) = 16.65 m/s.
Therefore, particle 1 has a velocity of 16.65 m/s in the positive x-direction at 3 seconds.
For particle 2, we know it has an acceleration of 7.45 m/s² in the negative y-direction. Since it starts from rest at the origin, we can again use the same equation of motion:
V₂ = V₀ + a₂t,
where V₂ is the velocity of particle 2 at time t, V₀ is the initial velocity (which is zero in this case), a₂ is the acceleration of particle 2, and t is the time (3 seconds in this case).
Using the values given, we substitute them into the equation:
V₂ = 0 + (-7.45 m/s²)(3 s) = -22.35 m/s.
Therefore, particle 2 has a velocity of -22.35 m/s in the negative y-direction at 3 seconds.
To find the speed of particle 1 with respect to particle 2, we need to find the magnitude of the relative velocity between the two particles. We can use the Pythagorean theorem:
Relative Speed = √((V₁ - V₂)²)
Plugging in the velocities we found:
Relative Speed = √((16.65 m/s - (-22.35 m/s))²)
Relative Speed = √((16.65 m/s + 22.35 m/s)²)
Relative Speed = √((39 m/s)²)
Relative Speed = √(1521 m²/s²) = 39 m/s.
Therefore, the speed of particle 1 with respect to particle 2 at 3 seconds is 39 m/s.