Uniform and accelerated motion: formulas, graphs

In Year 2 of high school, in a vocational school (istituto professionale), physics usually starts here: kinematics, that is, how to describe movement without yet asking what causes it. If it is October or November, the topic has just started or is about to, and the test usually falls five to eight weeks from now. There is time to understand it well, and it is worth doing so, because uniform and accelerated motion come back in the whole of the rest of the programme.

Here we look, with concrete examples (cars, trains, braking), at the two formulas you really need, how to read the graphs, the mistakes I see most often and three exercises worked out step by step. For recognising a proportionality from a graph there is a dedicated article; for how to write a measurement with its error, the one on measurement errors; for forces, which come next, the one on the inclined plane.

The quantities, in plain words

• Distance $s$: where the body is along the road, in metres (m).
• Time $t$: how much has passed, in seconds (s).
• Velocity $v$: how much distance is covered every second, in m/s.
• Acceleration $a$: by how much the velocity changes every second, in m/s². If $a=2{\text{m/s}}^2$, the velocity increases by 2 m/s every second.

Uniform motion in a straight line (MRU in Italian textbooks)

A car that always goes at 90 km/h on a straight road, a train at constant velocity: the velocity does not change. The distance grows steadily:

$s_0$ is where you are when the stopwatch starts (often $0$). From here you get the two versions you need almost every time:

Example: a train covers 120 km in an hour and a half. The velocity is $120/1.5=80$ km/h.

Distance-time graph (s-t)

In uniform motion it is a straight line. You read it like this:

• straight line sloping upwards: the body moves away; the steeper it is, the faster it goes;
• horizontal straight line: the body is at rest (the distance does not change);
• straight line sloping downwards: the body comes back (negative velocity).

The slope of the line is the velocity. If the graph is a curve, the velocity is changing.

Velocity-time graph (v-t)

In uniform motion it is a horizontal straight line: the velocity is always the same. The distance travelled is the area under the line (a rectangle: base times height, that is time times velocity).

Uniformly accelerated motion (MRUA in Italian textbooks)

When you set off from a traffic light, or brake before a bend, the velocity changes. If it changes by the same amount every second, the acceleration $a$ is constant and the motion is uniformly accelerated. There are two formulas:

$v_0$ is the velocity at the start (zero if you start from rest). When the problem does not give the time, you use a third formula, derived from the other two:

Graphs

In the v-t graph uniformly accelerated motion is a sloping straight line. The slope is the acceleration; the area under the line is the distance travelled (if you start from rest, a triangle: base times height divided by two).

In the s-t graph it is a curve (a parabola): the slope changes, so the velocity changes.

The mistakes that cost marks

Mixing km/h and m/s. This is the most frequent mistake. To go from km/h to m/s you divide by $3.6$: 72 km/h is $72/3.6=20$ m/s. The other way round you multiply: 10 m/s is 36 km/h. The acceleration formulas want m/s, so convert before calculating.

The sign of the acceleration. If a car brakes, the velocity decreases: choosing the direction of motion as positive, the acceleration is negative. If the problem says “brakes with a deceleration of 3 m/s²”, in the formulas you write $a=-3$.

Confusing average velocity and instantaneous velocity. The average is the total distance divided by the total time; the instantaneous one is what you read on the speedometer at that moment. A journey can have an average of 60 km/h without ever going at exactly 60 km/h.

Averaging velocities without thinking. For motion with constant acceleration, the average velocity between start and end is the arithmetic mean $(v_0+v)/2$. But if the motion changes type halfway (first it accelerates, then it goes at constant velocity), that shortcut does not hold: you calculate the total distance and the total time. You see it in the third exercise.

Reading the s-t graph as if it were a road. The curve is distance as a function of time, not the trajectory: a horizontal straight line is not a flat road, it is a body at rest.

Forgetting the units. An answer without units is worth little in a test: always write m, s, m/s or m/s².

Three worked exercises

1. Car at constant velocity

A car travels at $72$ km/h on a straight road. How many metres does it cover in $15$ s? How long does it take to cover $5$ km?

First I convert: $72/3.6=20$ m/s. In $15$ s:

For $5$ km, that is $5000$ m:

which is $4$ minutes and $10$ seconds ($250=4\cdot 60+10$). Check: at $72$ km/h you cover $72$ km in an hour, so $5$ km is about $1/14$ of an hour, that is about $4.2$ minutes. It works.

2. A car braking

A car travels at $54$ km/h and brakes with acceleration $a=-3{\text{m/s}}^2$ until it stops. Calculate the braking time and the distance covered.

I convert: $54/3.6=15$ m/s. At the end the velocity is zero, so from the first formula:

The braking distance:

Check with the formula that does not use time: $s=v_0^2/(2|a|)=225/6=37.5$ m, the same. A useful observation: if the velocity doubles (from 15 to 30 m/s) the braking distance becomes four times as large, because $v_0$ appears squared.

3. A start and then constant velocity

A train starts from rest with $a=0.5{\text{m/s}}^2$ for $40$ s, then keeps the velocity it has reached for another $60$ s. Calculate the velocity reached, the total distance and the average velocity over the whole journey.

First phase, accelerated. Velocity at the end:

Distance in this phase (it starts from rest, $v_0=0$):

Second phase, uniform at $20$ m/s for $60$ s:

Total distance $s=400+1200=1600$ m in $100$ s. The average velocity:

Note that $16$ m/s is not the arithmetic mean of $0$ and $20$ m/s (that would be $10$): in this case the shortcut does not hold because the motion changes type halfway. Check on the v-t graph: the area is a triangle ($\frac{1}{2}\cdot 40\cdot 20=400$) plus a rectangle ($60\cdot 20=1200$), total $1600$, as above.

How to prepare over the coming weeks

1. Learn to convert km/h and m/s without thinking: divide or multiply by $3.6$.
2. Redo the three exercises without looking, changing the numbers.
3. Draw by hand, one below the other, the s-t and v-t graphs of a car that sets off, goes at constant velocity and then brakes.
4. When the chapter is closed in class, try a timed test in the simulator for high school tests. It is not made for vocational schools and there is no test dedicated only to kinematics, but the physics questions show you where you stand with the method.

More material on the subject is on the physics page.

How we work in lessons

I give one-to-one tutoring online. On this topic I always start from a real situation (a braking car, a start) and from the graph drawn by hand, and only afterwards do I write the formula. Then we do exercises with your own numbers, especially the ones where the units or the sign hide the mistake.

If you want a hand

If the test is getting close and you want to see where you stand, you can book a free introductory call: we look together at where the gaps are and whether it makes sense to work on them.