Uniformly Accelerated Rectilinear Motion
You're riding a bus down the Periférico and the driver slams on the brakes because a car cut in front of him. Your body lurches forward: the velocity changed very quickly. How much space does a vehicle need to stop, and why is driving while looking at your phone so dangerous? With uniformly accelerated rectilinear motion (UARM) you'll be able to calculate it.
What UARM is
An electric scooter starts moving and its speedometer reads 0 km/h at the start, 4 km/h after the first second, 8 km/h after the second, and 12 km/h after the third. It gains speed at the same rate all the time. Does it cover the same distance each second? No: since it keeps going faster, each stretch is longer.
Speed is how far you travel per second (a positive number); velocity also has a sign, which indicates the direction along the line. Here we use velocity with a sign.
A car starts with an acceleration of $2\ \text{m/s}^2$. a) Explain in your own words what this piece of data means: by how much does its speed increase each second? b)…
A car starts with an acceleration of $2\ \text{m/s}^2$.
- a) Explain in your own words what this piece of data means: by how much does its speed increase each second?
- b) If it starts from rest, what speed will it have after $5\ \text{s}$?
A high-speed train accelerates from rest to $50\ \text{m/s}$ in $100\ \text{s}$. Calculate its acceleration and express the result in $\text{m/s}^2$.
A high-speed train accelerates from rest to $50\ \text{m/s}$ in $100\ \text{s}$. Calculate its acceleration and express the result in $\text{m/s}^2$.
A bicycle starts from rest and accelerates at $3\ \text{m/s}^2$ for $4\ \text{s}$. Calculate its final speed using $v = v_0 + a \times t$.
A bicycle starts from rest and accelerates at $3\ \text{m/s}^2$ for $4\ \text{s}$. Calculate its final speed using $v = v_0 + a \times t$.
The velocity equation
If you solve the definition of acceleration for $v$, you get the velocity equation. It makes sense: the initial velocity plus what you gain each second ($a$) times the seconds that pass ($t$).
Move the sliders for $v$ (initial velocity) and $a$: the graph of $v(t)$ is a slanted line, no longer horizontal as in uniform motion (UM), and the position $x(t)$ curves.
A motorcycle travels at $5\ \text{m/s}$ and accelerates at $2\ \text{m/s}^2$ for $6\ \text{s}$. What speed does it reach?
A motorcycle travels at $5\ \text{m/s}$ and accelerates at $2\ \text{m/s}^2$ for $6\ \text{s}$. What speed does it reach?
A car travelling at $20\ \text{m/s}$ brakes with an acceleration of $-4\ \text{m/s}^2$. a) Why is the acceleration negative? b) How long does it take to come to a…
A car travelling at $20\ \text{m/s}$ brakes with an acceleration of $-4\ \text{m/s}^2$.
- a) Why is the acceleration negative?
- b) How long does it take to come to a complete stop?
A bus reduces its speed from $25\ \text{m/s}$ to $15\ \text{m/s}$ in $5\ \text{s}$. a) Calculate its acceleration. b) Explain the meaning of the sign of the result.
A bus reduces its speed from $25\ \text{m/s}$ to $15\ \text{m/s}$ in $5\ \text{s}$.
- a) Calculate its acceleration.
- b) Explain the meaning of the sign of the result.
Watch the units: from km/h to m/s
Speedometers read km/h, but acceleration is given in $\text{m/s}^2$. Before calculating, convert everything to meters and seconds.
A car goes from $0$ to $90\ \text{km/h}$ in $10\ \text{s}$ with constant acceleration. a) Express the final speed in $\text{m/s}$. b) Calculate the acceleration. c)…
A car goes from $0$ to $90\ \text{km/h}$ in $10\ \text{s}$ with constant acceleration.
- a) Express the final speed in $\text{m/s}$.
- b) Calculate the acceleration.
- c) Calculate the distance covered while starting off.
A race car goes from $0$ to $108\ \text{km/h}$ in $3\ \text{s}$. a) Express the final speed in $\text{m/s}$. b) Calculate its acceleration. c) How many meters does it…
A race car goes from $0$ to $108\ \text{km/h}$ in $3\ \text{s}$.
- a) Express the final speed in $\text{m/s}$.
- b) Calculate its acceleration.
- c) How many meters does it cover in those $3\ \text{s}$?
The position equation
What distance does an accelerating object cover? $x = v \cdot t$ doesn't work, because the velocity changes. But we can use the average velocity.
A cyclist starts from rest and accelerates at $0.5\ \text{m/s}^2$ for $10\ \text{s}$. a) Calculate the speed reached. b) Calculate the distance covered using…
A cyclist starts from rest and accelerates at $0.5\ \text{m/s}^2$ for $10\ \text{s}$.
- a) Calculate the speed reached.
- b) Calculate the distance covered using $x = x_0 + v_0 \times t + \frac{1}{2} a \times t^2$.
A car travels at $10\ \text{m/s}$ and accelerates at $2\ \text{m/s}^2$ for $5\ \text{s}$. Calculate the distance it covers in that time.
A car travels at $10\ \text{m/s}$ and accelerates at $2\ \text{m/s}^2$ for $5\ \text{s}$. Calculate the distance it covers in that time.
A bicycle starts from rest at a traffic light and accelerates at $1.5\ \text{m/s}^2$ for $8\ \text{s}$. a) What is its speed at the end of the acceleration? b) What…
A bicycle starts from rest at a traffic light and accelerates at $1.5\ \text{m/s}^2$ for $8\ \text{s}$.
- a) What is its speed at the end of the acceleration?
- b) What distance has it covered in those $8\ \text{s}$?
Reading an equation and relating distance to velocity
There are two common situations. One: you're given the equation already written with numbers and you must identify $x_0$, $v_0$, and $a$ in it. The other: you aren't given the time, only velocities and acceleration.
Comparing with the general equation
In SI units, if $x = 3 + 2t + t^2$, you compare term by term with $x = x_0 + v_0 t + \frac{1}{2} a t^2$: $x_0 = 3\ \text{m}$, $v_0 = 2\ \text{m/s}$, and $\frac{1}{2}a = 1$, so $a = 2\ \text{m/s}^2$. Careful: the coefficient of $t^2$ is half of $a$.
Without knowing the time
From $t = \dfrac{v - v_0}{a}$ and $x - x_0 = \dfrac{v_0 + v}{2}\, t$ we get:
The position of an object, in International System units, is given by $x = 3 + 2 \times t + t^2$. By comparing with $x = x_0 + v_0 \times t + \frac{1}{2} a \times t^2$:…
The position of an object, in International System units, is given by $x = 3 + 2 \times t + t^2$. By comparing with $x = x_0 + v_0 \times t + \frac{1}{2} a \times t^2$:
- a) Determine its initial position $x_0$, its initial velocity $v_0$ and its acceleration $a$.
- b) Calculate its velocity when $t = 4\ \text{s}$.
- c) Calculate its position at that same instant.
A skateboarder goes down a ramp: they enter at $2\ \text{m/s}$ and accelerate at $3\ \text{m/s}^2$. The scene is shown in the figure. a) How long does it take to reach…
A skateboarder goes down a ramp: they enter at $2\ \text{m/s}$ and accelerate at $3\ \text{m/s}^2$. The scene is shown in the figure.
- a) How long does it take to reach $20\ \text{m/s}$?
- b) What distance do they cover in that time?
Stages: the area under the v-t graph
A Metro train doesn't accelerate the whole time: it starts off, then keeps its velocity. For trips like this, the motion is split into stages and each one is calculated separately. The $v$-$t$ graph helps: the area under the line is the total distance.
An underground train starts from rest with an acceleration of $1.2\ \text{m/s}^2$ for $15\ \text{s}$. a) Calculate the speed it reaches and express it in $\text{km/h}$…
An underground train starts from rest with an acceleration of $1.2\ \text{m/s}^2$ for $15\ \text{s}$.
- a) Calculate the speed it reaches and express it in $\text{km/h}$ as well.
- b) Calculate the distance covered in that time.
A train starts from rest and accelerates at $0.5\ \text{m/s}^2$ for $60\ \text{s}$; then it maintains the speed reached for $300\ \text{s}$. The $v$-$t$ graph for the…
A train starts from rest and accelerates at $0.5\ \text{m/s}^2$ for $60\ \text{s}$; then it maintains the speed reached for $300\ \text{s}$. The $v$-$t$ graph for the journey is shown in the figure.
- a) What speed does it reach at the end of the acceleration?
- b) What distance does it cover in each of the two stages?
- c) What is the total distance?
- d) Calculate the average speed for the whole journey, in $\text{m/s}$.
A real braking situation: reaction time
You're riding your bike at 20 km/h and suddenly a dog runs out 10 m ahead. Before you squeeze the brakes, a few tenths of a second go by in which you keep going at full speed. That's why the distance needed to stop doesn't depend only on the brakes.
A car travelling at $72\ \text{km/h}$ brakes with an acceleration of $-5\ \text{m/s}^2$ until it stops. The $v$-$t$ graph of the braking is shown in the figure. a)…
A car travelling at $72\ \text{km/h}$ brakes with an acceleration of $-5\ \text{m/s}^2$ until it stops. The $v$-$t$ graph of the braking is shown in the figure.
- a) Express the initial speed in $\text{m/s}$.
- b) How long does the braking last?
- c) What distance does it cover while braking?
A tram travelling at $54\ \text{km/h}$ brakes with an acceleration of $-1.5\ \text{m/s}^2$ until coming to a stop at the station. a) Express the initial speed in…
A tram travelling at $54\ \text{km/h}$ brakes with an acceleration of $-1.5\ \text{m/s}^2$ until coming to a stop at the station.
- a) Express the initial speed in $\text{m/s}$.
- b) Calculate the braking time.
- c) Calculate the distance covered during braking.
A car travels at $90\ \text{km/h}$ when the driver sees an obstacle. It takes $0.8\ \text{s}$ to react (reaction time) and then brakes with $a = -6.25\ \text{m/s}^2$.…
A car travels at $90\ \text{km/h}$ when the driver sees an obstacle. It takes $0.8\ \text{s}$ to react (reaction time) and then brakes with $a = -6.25\ \text{m/s}^2$. The two stages of stopping are shown in the figure.
- a) Express the speed in $\text{m/s}$ and calculate the distance covered during the reaction time.
- b) Calculate the braking time and the braking distance.
- c) What is the total stopping distance from the moment the driver sees the obstacle?
- d) Based on the results, explain why using a cell phone while driving is so dangerous.
Braking distance and road safety
Braking until stopping with constant $a$: $t = \dfrac{v_0}{|a|}$ and $d = \dfrac{v_0^2}{2|a|}$. There are two important consequences.
Why? When you double the initial velocity, the braking time doubles and the average velocity during braking also doubles. Distance = average velocity × time, so it is multiplied by $2 \times 2 = 4$.
A train travelling at $30\ \text{m/s}$ must stop at the next station by braking with an acceleration of $-0.5\ \text{m/s}^2$. a) How long will the braking last? b) What…
A train travelling at $30\ \text{m/s}$ must stop at the next station by braking with an acceleration of $-0.5\ \text{m/s}^2$.
- a) How long will the braking last?
- b) What distance will it cover while braking?
- c) At what distance from the station must the train driver start braking?
- d) Explain why this distance is so large even though the acceleration is small.
Two cars travel at $30\ \text{m/s}$ and perform an emergency stop: the first with $a = -5\ \text{m/s}^2$ and the second with $a = -6\ \text{m/s}^2$. a) Calculate the…
Two cars travel at $30\ \text{m/s}$ and perform an emergency stop: the first with $a = -5\ \text{m/s}^2$ and the second with $a = -6\ \text{m/s}^2$.
- a) Calculate the braking time and braking distance for each car.
- b) If there is an obstacle $80\ \text{m}$ from the point where they start braking, which of the two hits it? Justify your answer with calculations.
- c) Explain what quality of a car the braking distance describes and why it matters in road safety.
We want to test whether "doubling the speed doubles the braking distance" is true. Consider a car that brakes with $a = -5\ \text{m/s}^2$. a) Calculate the braking time…
We want to test whether "doubling the speed doubles the braking distance" is true. Consider a car that brakes with $a = -5\ \text{m/s}^2$.
- a) Calculate the braking time and braking distance if it travels at $20\ \text{m/s}$.
- b) Repeat the calculation if it travels at $40\ \text{m/s}$.
- c) How many times larger is the braking distance when the speed is doubled? Was the statement true?
- d) Explain the result, taking into account that doubling the speed doubles both the braking time and the average speed during braking.
Chases and catch-ups (extension)
A car moving at constant speed passes a parked police car, which then starts moving to catch up. When does it catch up? When both are at the same position: the position equations are set equal.
A car travels at a constant speed of $20\ \text{m/s}$ and, as it passes a stationary police car, the police car starts off and pursues it with a constant acceleration of…
A car travels at a constant speed of $20\ \text{m/s}$ and, as it passes a stationary police car, the police car starts off and pursues it with a constant acceleration of $2\ \text{m/s}^2$.
- a) Write the position equations for both cars, taking the overtaking point as the origin.
- b) Calculate the instant at which the police car catches up with the other car. You will obtain two mathematical solutions: explain why one of them is discarded.
- c) At what distance from the origin does it catch up?
- d) Calculate the speed of the police car at that instant and compare it with that of the other car.
A cyclist travels at $10\ \text{m/s}$ and, at the moment they pass a stationary moped, the moped starts off and pursues them with an acceleration of $4\ \text{m/s}^2$.…
A cyclist travels at $10\ \text{m/s}$ and, at the moment they pass a stationary moped, the moped starts off and pursues them with an acceleration of $4\ \text{m/s}^2$.
- a) Write the position equations for both vehicles.
- b) Solve the resulting quadratic equation to find the time when the moped catches up.
- c) At what distance from the starting point does this occur?
- d) What speed is the moped travelling at that moment?
In a training session, runner A passes the starting line at a constant speed of $6\ \text{m/s}$ at the exact instant runner B starts from rest with…
In a training session, runner A passes the starting line at a constant speed of $6\ \text{m/s}$ at the exact instant runner B starts from rest with $a = 1\ \text{m/s}^2$.
- a) Who is ahead when $t = 8\ \text{s}$? Calculate the position of each runner.
- b) Calculate the instant when B catches up with A, and the position of the overtake.
- c) What speed is B travelling at that instant?
- d) Explain why B, despite starting off slower, ends up catching up with A.
Back to the start: the bus that brakes suddenly needs $d = v_0^2 / (2|a|)$ in addition to what it covers during the reaction time. Looking at your phone lengthens that time and adds uncontrolled meters.