Connect vectors with motion and work

Connect vectors with motion and work

Identify what each component measures and keep its units. Velocity and force are vectors, while speed and work are scalars. Use the model assumptions to decide which calculation fits the physical question.

Appendix B uses three physics settings to motivate three-dimensional vectors. Its syllabus references describe the supplied 2025 notes.

Particle motion in 3D. Constant-velocity and constant-acceleration models simplify motion. A boomerang can change velocity and acceleration in multiple directions. The source illustrates a projectile under constant downward acceleration with its linked three-dimensional projectile demonstration.

Collisions. Straight-line particle collisions are a one-dimensional simplification. A baseball meeting a bat and repeated billiard-ball collisions involve more directions. The source links a collision model for two balls of equal volume and mass.

Work done by a force. For a constant force $\mathbf F$ and displacement $\mathbf s$, work is displacement length times the force component along that displacement:

$$W=|\mathbf s|(|\mathbf F|\cos\theta)=\mathbf F\cdot\mathbf s.$$

The sign matters: a force component opposing the displacement does negative work. The two source QR references are vectors and work and the mathematics of work. These are the original external resource addresses; the chapter’s explanation does not depend on opening them.

Distinguish a highest point from coming to rest

A particle has position $\mathbf r(t)=(2t,6t-t^2,0)$ metres for $0\le t\le6$ seconds, with the second coordinate representing height. Find its velocity and acceleration. Locate its highest point and decide whether it is at rest there. State which component controls the height conclusion.

After your attempt: compare the reasoning

Differentiating components gives $\mathbf v(t)=(2,6-2t,0)$ metres per second and $\mathbf a(t)=(0,-2,0)$ metres per second squared. Height is $6t-t^2=9-(t-3)^2$, so the highest point is $(6,9,0)$ at $t=3$. There the vertical velocity is zero, but $\mathbf v(3)=(2,0,0)$ and speed is 2 metres per second. Zero vertical velocity does not mean zero velocity. The downward vertical acceleration also confirms that the stationary height is a maximum.