Tracking a Particle's Motion in the xy Plane

[neutron star]

Homework Statement

A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 3.0 m, y = 6.0 m, and has velocity v = 1.0 m/s i-hat + 7.0 m/s j-hat. The acceleration is given by the vector a = 9.0 m/s$$^2$$ i-hat + -1 m/s$$^2$$ j-hat.


What is the velocity vector at t=3.0s

What is the position vector at t=5.0s

What is the magnitude and direction of the position vector

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

Hi neutron star!

The standard constant acceleration equations apply to the the components (of position velocity and acceleration) in any particular direction …

so try them in the 9i - j direction (and of course, there's zero acceleration in the perpendicular direction).

 

[tomcue]

I would approach this problem by first understanding the given information. We are tracking a particle's motion in the xy plane, which means we are looking at its position and velocity in the horizontal and vertical directions. The particle has a constant acceleration, which means its velocity is changing at a constant rate. We are given the initial position and velocity of the particle, as well as its acceleration vector.

To find the velocity vector at t=3.0s, we can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the given values, we get:

v = (1.0 m/s i-hat + 7.0 m/s j-hat) + (9.0 m/s^2 i-hat + -1 m/s^2 j-hat) * 3.0s

= 28.0 m/s i-hat + 4.0 m/s j-hat

Therefore, the velocity vector at t=3.0s is 28.0 m/s i-hat + 4.0 m/s j-hat.

To find the position vector at t=5.0s, we can use the formula s = ut + 1/2at^2, where s is the final position, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the given values, we get:

s = (3.0 m i-hat + 6.0 m j-hat) + (1.0 m/s i-hat + 7.0 m/s j-hat) * 5.0s + 1/2 * (9.0 m/s^2 i-hat + -1 m/s^2 j-hat) * (5.0s)^2

= 3.0 m i-hat + 6.0 m j-hat + 5.0 m i-hat + 35.0 m j-hat + 112.5 m i-hat + -6.25 m j-hat

= 120.5 m i-hat + 35.75 m j-hat

Therefore, the position vector at t=5.0s is 120.5 m i-hat + 35.75 m j-hat.