Connected ideas

Motion graphs and calculus

Connect velocity, gradient and signed area with movement.

Predict, test and explain

Step 1 of 3

Predict

Start at 4 m/s with acceleration −2 m/s² for 4 seconds. What is the displacement?

v(t)=−2t+4v(t)=-2t + 4
00
11
22
33
44
55
66
−8-8
−4-4
00
44
88
t  (s)t\;(\text{s})
v  (m/s)v\;(\text{m/s})

Scroll the diagram sideways to see every label.

v(T)=−4  m/sv(T)=-4\;\text{m/s}Displacements(T)=0  ms(T)=0\;\text{m}

The gradient is acceleration. Area above the time axis is positive displacement; area below is negative. Distance adds the magnitudes of both areas. Displacement starts at zero.

Check the working
QuantityValue
Elapsed time4 s
Velocity-4 m/s
Acceleration-2 m/s²
Signed area / displacement∫04(−2t+4) dt=0  m\int_0^{4}(-2t + 4)\,dt=0\;\text{m}
Total distance travelled8 m
Direction change before T2 s

InvestigateCan displacement be zero while distance is positive?

Explore and compare

Choose the investigation above. Predict what will change before you move a control.

Try it in the simulation. Change one thing at a time.

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Hints, self-check and connections
Try a self-check

Fixed values, separate from the diagram controls.

Velocity is v(t) = −2 m/s for 3 seconds. Find displacement and total distance.

v(t)=−2,0≤t≤3v(t)=-2,\quad 0\leq t\leq3

Hint

The velocity graph lies below the time axis. Distinguish signed area from its magnitude.

Reveal answer

The signed rectangular area is negative six metres. The object travels six metres in the negative direction.

s=−6  m,d=6  ms=-6\;\text{m},\quad d=6\;\text{m}

Build an explanation

Hint 1

The gradient of a velocity–time graph is acceleration. The gradient of a displacement–time graph is velocity.

Hint 2

Signed area under velocity gives displacement. Below-axis areas subtract.

Hint 3

Total distance adds the magnitudes of the areas. Split at a direction change where velocity crosses zero.

Worked example

This example uses fixed values, separate from the diagram controls.

An object starts at velocity 4 m/s and has constant acceleration −2 m/s² for 4 seconds. Find its displacement and distance.

  1. Velocity is v(t) = 4 − 2t, so it reaches zero at t = 2 seconds.
  2. The positive area from 0 to 2 seconds is ½ × 2 × 4 = 4 m.
  3. The negative area from 2 to 4 seconds is −4 m.
  4. Displacement is 4 − 4 = 0 m; total distance is 4 + 4 = 8 m.

Answer: Displacement: 0 m. Distance: 8 m.

Connect this idea

Watch out: Negative velocity means travel in the opposite direction. Displacement can be negative or zero, while total distance remains nonnegative. This model starts with zero displacement.