Four forces act on a sprinter. Balance them and the sprinter keeps a constant velocity. Unbalance them and the net force makes the sprinter speed up or slow down. Push the sliders and watch.
Model: horizontal forces averaged over a stride. Air resistance = 0.24 × speed² (still air). Vertically, reaction equals weight because the sprinter is not accelerating up or down.
Friction here is the average over a whole stride. At each foot contact the forward push is much larger, but the foot is only on the ground for part of the stride.
Weight acts down from the centre of mass and reaction acts up from the point of contact. Friction acts forwards at the foot, and air resistance acts backwards from the centre of mass. Arrow length shows size: compare friction with air resistance to find the net force.
Linear motion in one line. Linear motion is motion in a straight or curved line, with all body parts moving the same distance in the same direction over the same time. Velocity is the rate of change of displacement (m/s); acceleration is the rate of change of velocity (m/s²). Distance–time and velocity–time graphs have their own lab: Motion Graphs Lab.
| Phase of 100 m | Horizontal forces | Net force | Effect |
|---|---|---|---|
| Start | Friction much larger than air resistance | Forwards | Acceleration |
| Middle | Friction equals air resistance | Zero (balanced) | Constant velocity |
| End | Air resistance larger than friction | Backwards | Deceleration |
In every phase the vertical forces are balanced: reaction equals weight, so draw them the same length. (Real forces change within each stride; exam diagrams show the average.)
Read the scenario, then build the diagram. Choose how reaction compares with weight, how friction compares with air resistance, and which way friction acts. The diagram redraws as you choose. Then check your marks.
Scenario 1/5Marks 0/0
Make a choice in each row, then check. Weight is always drawn the same length; your choices set the other arrows.
Each card describes a moment in sport. Which of Newton's laws explains it best?
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Pick a law.
Two sprinters push off the blocks. Athlete A's mass and force are fixed. Set athlete B's force so both have the same acceleration and finish the first 10 m together. Remember a = F ÷ m: a heavier athlete needs a bigger force for the same acceleration.
Work it out first: a = F ÷ m for A, then F = m × a for B.
Game model: a constant average force over the first 10 m, ignoring air resistance (small at low speed). Real sprinters' forces fall as they speed up.
Static friction is the maximum friction force that can be applied before a body starts to slide. Sprinters want high static friction so their spikes grip and the foot does not slip as they push back.
| Law | Sprinter at the start | Swimmer at a turn |
|---|---|---|
| 1st | Stays at rest in the blocks until an external force acts (the reaction force from the blocks when the legs push back on them) | Would keep gliding at the same velocity unless an external force (water resistance) acts |
| 2nd | The bigger the force on the blocks, the greater the acceleration, in the direction of the force | A harder push off the wall gives a greater acceleration away from it |
| 3rd | Pushes back on the blocks (action); the blocks push forwards on the sprinter (reaction) | Pushes back on the wall (action); the wall pushes forwards on the swimmer (reaction) |
State Newton's three laws of motion.
A 72 kg sprinter experiences a forward friction force of 310 N and an air resistance force of 22 N. Calculate the net force and the sprinter's acceleration. Show your working.
Draw and label a free-body diagram for a sprinter running at a constant velocity in the middle of a 100 m race.
Use Newton's third law to explain how a swimmer pushes off the wall at a turn.
Explain two factors that affect the friction between a footballer's boots and the pitch.
Using all three of Newton's laws of motion, explain how a footballer takes a penalty kick.