Name the system and the requested quantity
A motion question, a force question and an energy question can describe the same object while requiring different models. Write what the question asks you to find, then list the given quantities with units. A sketch of the path or of the forces can reveal which relationship is relevant.
For introductory mechanics, keep any stated assumptions visible: a constant mass, constant acceleration or negligible resistance. If a formula depends on one of these assumptions, it belongs in the explanation. A correct calculation using an unsuitable model does not answer the physical problem.
Distinguish average speed from average velocity
Average speed is total path length divided by elapsed time. Average velocity is displacement divided by elapsed time, including direction. A runner who travels 100 m out and 100 m back in 50 s has average speed 200/50 = 4 m/s, but average velocity zero because the final position equals the initial position.
The cyclist example gives an average speed of 5 m/s from 150 m in 30 s. These totals do not establish the speed at a specific instant. If the question asks for instantaneous speed, you need additional information, such as a local gradient on an appropriate graph.
Read the graph axes before using its slope
On a graph of cumulative distance traveled against time, the slope gives speed. A straight segment from (10 s, 20 m) to (30 s, 80 m) has slope (80 − 20)/(30 − 10) = 3 m/s. Use changes between the endpoints, not 80/30, because this segment begins after the origin.
A position-time graph uses position rather than cumulative distance; its slope gives velocity, which can be negative. A velocity-time graph has acceleration as its slope, while its signed area over an interval gives displacement. The same rising line has different meanings when the vertical quantity changes.
Use net force in Newton’s second law
For a constant-mass object in an inertial frame, the vector sum of external forces is Fnet = ma. Suppose a 2 kg cart has a 10 N horizontal force right and a 4 N opposing force left. Taking right as positive, Fnet = 10 − 4 = 6 N, so acceleration is 6/2 = 3 m/s² right.
The 10 N applied force alone is not the net force. A force diagram should include forces on the chosen object and the relevant directions. Vertical forces may balance while horizontal forces do not. State the axis convention before calculating signed components.
Connect work with energy under stated conditions
For a constant force parallel to displacement, work is force times displacement. A 6 N force acting through 3 m in the same direction does 18 J of work. If the force points at an angle, its component along the displacement matters; the general constant-force relationship is W = Fd cos θ.
Near Earth’s surface, a change in gravitational potential energy can be modeled as ΔU = mgΔh when g is effectively constant. Raising 2 kg by 1.5 m with g = 9.8 m/s² gives 29.4 J. Name the height change and the model before comparing energy values.
Carry units through conversions and checks
To convert 54 km/h to m/s, multiply by 1000 m/km and divide by 3600 s/h: 54 × 1000/3600 = 15 m/s. Unit cancellation shows why the factors belong in that order. A force divided by mass produces acceleration because N/kg is equivalent to m/s².
Check the requested quantity, signs, units and plausible scale independently. A dimensional check can reveal a wrong formula but cannot prove that the chosen model fits the situation. Bring your diagram and attempted setup to Lirno when the uncertainty is in the model rather than the arithmetic.
Use an energy account when several forces do work
A constant 12 N horizontal push moves a box 3 m to the right. A constant 4 N friction force opposes the motion. The push does +36 J of work and friction does −12 J. If these are the only forces doing work, net work is +24 J, so the box’s kinetic energy increases by 24 J.
For a box of mass 2 kg starting at rest, ΔK = ½mv² gives 24 = ½ × 2 × v², hence v = √24 ≈ 4.90 m/s. A force check gives net force 8 N and acceleration 4 m/s²; v² = 2as = 24 agrees under the same constant-force conditions. Using the push alone would miss the energy transferred by friction.
