Uniformly Accelerated Rectilinear Motion (UARM)
A car that starts moving at a traffic light on the Autopista Duarte doesn't keep its speed: it gains it little by little, always at the same rate. One that brakes hard loses speed just as fast every second. That motion is UARM (uniformly accelerated rectilinear motion), and it is the basis of road safety: stopping distances, reaction times, and chases. Here you will answer a key question: why can't a car stop "dead" even when the brakes are good?
What UARM is
A motoconcho (motorcycle taxi) starts moving and its speedometer reads 0 km/h at the start, 4 km/h at the first second, 8 km/h at the second, and 12 km/h at the third: it gains speed always at the same rate. Does it cover the same distance each second, or more and more?
It covers more and more distance per second, because its velocity keeps growing.
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 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
It also works for finding the acceleration: $a = \dfrac{v - v_0}{t}$. If the result is negative, the object is braking.
Move the sliders for $v$ (initial velocity) and $a$: the $v(t)$ graph is a slanted line, not horizontal as in UR M, 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.
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 $\text{m/s}^2$.
The position equation
You can also read the data from a given equation. If $x = 3 + 2t + t^2$, comparing with the general formula: $x_0 = 3\ \text{m}$, $v_0 = 2\ \text{m/s}$, and $\frac{1}{2}a = 1$, that is, $a = 2\ \text{m/s}^2$.
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 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 $3\ \text{s}$?
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 $8\ \text{s}$?
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.
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?
Braking and combined stages
You are driving on the Autopista Duarte and suddenly a dog crosses the road. Before you step on the brake, some time passes during which you are still going at full speed. Why does the distance needed to stop not depend only on the brakes?
When a motion has several phases (for example, a train that accelerates and then goes at constant velocity), each phase is calculated separately: the final velocity of one is the initial velocity of the next. The average velocity of the whole trip is the total distance divided by the total time.
Go back to the beginning: the car doesn't stop dead because it first covers the reaction distance at constant velocity and then the braking distance, which grows with the square of the velocity.
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.
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 $\text{m/s}$.
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.
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.
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 with acceleration
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.