Question

An object's position in a given coordinate system is described by the vector r = t²i - (3t + 3)j. Assume the object moves without air resistance. What is the magnitude of the object's velocity between 1 and 3 seconds?

Answer 1

The magnitude of the object's velocity between 1 and 3 seconds is approximately 3.61 m/s at t=1 s and 6.71 m/s at t=3 s.

Step-by-step explanation:

The object's position is described by the vector r = t²i - (3t + 3)j. To find the object's velocity, we need to take the derivative of the position vector with respect to time. In this case, the object's velocity is given by v = d(r)/dt = 2ti - 3j. To find the magnitude of the object's velocity between 1 and 3 seconds, we substitute the values of t into the velocity equation and calculate the magnitude using the Pythagorean theorem.

At t = 1 second, the magnitude of the object's velocity is |v| = sqrt((2*1)^2 + (-3)^2) = sqrt(4 + 9) = sqrt(13) approximately 3.61 m/s. At t = 3 seconds, the magnitude of the object's velocity is |v| = sqrt((2*3)^2 + (-3)^2) = sqrt(36 + 9) = sqrt(45) approximately 6.71 m/s.