Question

If the vector components of the position of a particle moving in the xy plane as a function of time are x = (2.7 m/s2)t2i and y = (5.1 m/s3)t3j, at what time t is the angle between the particle's velocity and the x axis equal to 45°?

Answer 1

Answer:

Time: 0.35 s

Explanation:

The position vector of the particle is

$r=(2.7m/s^2)t^2 i+(5.1 m/s^3)t^3j$

where the first term is the x-component and the 2nd term is the y-component.

The particle's velocity vector is given by the derivative of the position vector, so:

$v=r'(t)=(2.7\cdot 2)t i + (5.1\cdot 3)t^2 j=5.4t i+15.3t^2j$

The particle's velocity has an angle with the x-axis of 45 degrees when the x and the y component have same magnitude. Therefore:

$5.4t=15.3t^2\\5.4=15.3t\\t=\frac{5.4}{15.3}=0.35 s$