An object has a velocity of (0.4 x + -1.7 y) m/s at the origin at t = 0. It experiences a constant acceleration of (0.2 x + -1 y) m/s2. What is the distance between the object and the origin at t = 3 s?
To find the distance between the object and the origin at t = 3 s, we need to integrate the velocity function to obtain the position function.
The velocity of the object is given by:
v(t) = (0.4x - 1.7y) m/s
To integrate this velocity function with respect to time, we need to treat x and y as constants since there are no time-dependent terms involving them. So the integration can be done separately for the x and y components:
∫v(t) dt = ∫(0.4x - 1.7y) dt
Integrating with respect to time gives us the position function:
r(t) = (0.4x - 1.7y)t + C
Here, C is the constant of integration, which we'll determine using the initial conditions provided.
At t = 0, the object is at the origin (0,0), so we can substitute these values into the position function:
r(0) = (0.4(0) - 1.7(0))(0) + C
r(0) = C
Therefore, the value of C is 0.
Substituting the given time t = 3 s into the position function gives us:
r(3) = (0.4x - 1.7y)(3)
So, the distance between the object and the origin at t = 3 s is given by the magnitude of the position vector:
| r(3) | = sqrt((0.4x - 1.7y)(3))^2 + (0.4x - 1.7y)(3))^2)
Simplifying further, we have:
| r(3) | = sqrt((0.12x - 5.1y)^2 + (0.12x - 5.1y)^2)
Therefore, the distance between the object and the origin at t = 3 s is sqrt((0.12x - 5.1y)^2 + (0.12x - 5.1y)^2).