At t=0 a particle leaves the origin with a velocity of 5.0 m/s in the positive y direction, its acceleration is given by a=3i-2j m/s^2
At the instant the particle reaches its maximum y coordinate, how far is it from the origin?
I'm not sure how to do a question like this, do you have to integrate it or how do you solve this?
To find the distance of the particle from the origin when it reaches its maximum y coordinate, you need to analyze the motion of the particle in the y-direction.
Since the acceleration of the particle is given by a=3i-2j m/s^2, we first need to find the equation for the particle's velocity in the y-direction.
v = ∫a dt, where ∫ indicates integration and t is the time elapsed.
Integrating the acceleration function, we get:
∫(3i-2j) dt = 3∫i dt - 2∫j dt
Since j represents motion in the y-direction, we only need to integrate -2j dt.
∫j dt = -2tj
Therefore, the velocity in the y-direction is:
v = 3∫i dt - 2∫j dt = 3∫i dt - 2(-2tj) = 3ti + 4tj
Now, we can find the equation for the particle's position in the y-direction by integrating the velocity function:
r = ∫v dt
∫(3ti + 4tj) dt = 3∫ti dt + 4∫tj dt
Since we are interested in the maximum y coordinate, we can set the initial position at t = 0 as y = 0.
∫ti dt = (1/2)ti^2, ∫tj dt = (1/2)tj^2
So, the position equation becomes:
r = (3/2)ti^2 + (2/2)tj^2 = (3/2)ti^2 + tj^2
To find the maximum y-coordinate, we need to differentiate this equation with respect to t, set it equal to zero, and solve for t.
dy/dt = 0
d/dt((3/2)ti^2 + tj^2) = 0
Integrating both terms separately, we get:
3ti = C₁, where C₁ is a constant of integration
tj = C₂, where C₂ is a constant of integration
By substituting these into the original equation, we can solve for the constants:
(3/2)C₂^2 + C₁ = 0
From this equation, we can solve for t and find the maximum y-coordinate, and finally, calculate the distance of the particle from the origin using the Pythagorean theorem:
Distance = √[(position in x-axis)^2 + (position in y-axis)^2]