Problem statementSolution video |

**DISCUSSION THREAD**

You can ask questions or answer questions of others here. You can learn from either.

**DISCUSSION**

Shown in the animation below is the path taken by Particle P.

- As expected the velocity vector is always tangent to the path since
=**v***v*.**e**_{t} - The acceleration vector, in general, has both tangential and normal components of acceleration. The normal component always points in the same direction as the unit normal vector,
. The acceleration vector either points "forward" of the motion (for which the rate of change of speed is positive; i.e., increasing is speed), or points "backward" of the motion (for which the rate of change of speed is negative; i.e., slowing down). In the animation below, can you identify from the angle between the velocity and acceleration vectors whether P is increasing or decreasing in speed?**e**_{n} - The rate of change of speed is found from projecting the acceleration vector onto the unit tangent vector: v_dot = a •
, where**e**_{t}=**e**_{t }/*v**v.* - The radius of curvature can be found from the magnitude of the acceleration vector.

How are you guys plotting the acceleration and velocity vectors? Do you plot the position and draw the vectors at t=10?

Yeah, I didn't think we needed to plot out the path so I calculated the cartesian coordinates of P at t=10s and wrote that next to a point from which I drew Ap and Vp at the magnitude/direction they would be at t=10s.

If the animation shows the position curve, why does it go up toward y+ when the y component of position is -3t^2? Shouldn't it go down toward y-?

ENRICO: Good catch. The path of the point should be moving downward for large time t. I had an error in my simulation model: instead of prescribing a y-component of the motion on the particle, I had incorrectly prescribed a y-component of force.

Thank you for pointing out this error. I have fixed the simulation and have re-posted the animation of the simulation.