Uniformly accelerated rectilinear motion
A car pulling away from traffic lights does not keep a constant speed: it gains speed little by little, always at the same rate. One that brakes hard loses speed just as quickly every second. This motion, in a straight line and with constant acceleration, is UARM.
What UARM is
If you measure the position of a ball rolling down a ramp every second ($1\ \text{s} \to 0{,}5\ \text{m}$; $2\ \text{s} \to 2\ \text{m}$; $3\ \text{s} \to 4{,}5\ \text{m}$), you see that the distances covered in each interval keep growing: what mathematical laws describe this motion, in which the velocity changes at a constant rate?
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 $5\ \text{s}$?
The velocity equation
Move the sliders for $v$ (initial velocity) and $a$: the $v(t)$ graph is a sloping straight line (no longer horizontal as in constant-velocity motion) and the position $x(t)$ curves.
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 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.
The position equation
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.
A real braking situation: reaction time
If you are driving at $90\ \text{km/h}$ and something unexpected appears on the road, your foot does not hit the brake instantly: several crucial tenths of a second go by while the car keeps moving at full speed. How do you calculate the total distance you travel from the moment you see the obstacle until the car comes to a complete stop?
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 mobile phone while driving is so dangerous.
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?