Physics › Mechanics › Motion graphs and the SUVAT equations
Motion graphs and the SUVAT equations
Motion graphs encode everything through two operations, gradient and area, and they work for any motion at all. The four SUVAT equations are faster, but they carry one condition, and ignoring it produces confident wrong answers.
IN THIS TOPIC
- Use v = Δs/Δt and a = Δv/Δt, distinguishing average from instantaneous values.
- Read gradients and areas from motion graphs for uniform and non-uniform acceleration.
- Select and apply the right constant-acceleration equation, and recognise when none of them is valid.
WHAT YOU PROBABLY THINK
SUVAT always works.
The quantities and their definitions
Motion runs on four quantities: displacement s, the vector version of distance; velocity, the rate of change of displacement; and acceleration, the rate of change of velocity. The defining equations are on the data sheet:
Applied over an interval these give average values; shrink the interval and they become instantaneous ones, which is exactly what the gradient of a graph reads off at a point.
Reading the graphs
Every motion graph is decoded with the same two operations: gradient and area. On a displacement-time graph, the gradient is the velocity; a curve bending upward means the object is speeding up, and the gradient of the tangent gives the instantaneous velocity at that moment.
On a velocity-time graph, the gradient is the acceleration and the area beneath is the displacement. The area is the more valuable half, because it works for any shape of graph, including motions no equation describes. On an acceleration-time graph, the area gives the change in velocity.
Signs carry meaning throughout. Negative velocity is motion the other way, and area below the time axis subtracts from the displacement, which is how a ball thrown up and caught again ends with zero displacement despite plenty of distance travelled.
The bouncing ball is the specification's own example, and it rewards a careful look. During every flight, up or down, the only force is gravity, so every sloped section has the same gradient, −g. The bounce itself is the near-vertical jump, and each jump is a little shorter than the last as the ball loses speed.
The four equations, and the condition
For constant acceleration only, four equations connect the five quantities s, u, v, a, t:
Each equation omits exactly one of the five quantities, which is how you choose between them.
WORKED EXAMPLE
Choosing the equation
A car brakes from 24 m s−1 to rest over 60 m. Find the deceleration.
List what is known: u = 24, v = 0, s = 60. Wanted: a. The quantity not involved is t, so pick the equation without t: v2 = u2 + 2as.
0 = 242 + 2a × 60, so a = −576/120 = −4.8 m s−2. The minus sign is the deceleration doing its job.
The condition is the part people skip. Acceleration must be constant. A falling object with air resistance does not have constant acceleration, so SUVAT is simply invalid there: that is a velocity-time graph problem, and the displacement comes from the area. Free fall without air resistance qualifies, with a = g downward.
THE EXAM BIT
- Before touching SUVAT, ask whether the acceleration is constant. Air resistance, engines easing off, curved velocity-time graphs: all of them disqualify the equations, and questions are built to catch it.
- Write the SUVAT list, fill in knowns, mark the wanted quantity, and pick the equation missing the one you neither have nor need. It is faster than remembering which equation “does” which problem.
- Set a sign convention in the first line and keep it. An object thrown upward with downward g positive is where most sign errors are born.
- “Instantaneous” means the gradient of the tangent at a point; “average” means total change over total time. Questions ask for the difference in words and expect it in method.
- On any graph question, say what the gradient and the area mean for those axes before calculating. That sentence is often a mark by itself.
CHECK YOURSELF
A car brakes uniformly from 24 m s−1 to rest over 60 m. How long does the braking take?
Show a hint
You now know u, v, s and a. The quickest route uses the equation without a.
Show the answer
Using s = 12(u + v)t: 60 = 12 × t, so t = 5.0 s.
The same answer follows from v = u + at with a = −4.8 m s−2 from the previous part; agreement between routes is the free check the four equations offer.
Gradient and area read every graph.
SUVAT needs constant acceleration.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.