Uniformly Accelerated Rectilinear Motion (UARM)
A Metrobús bus pulls away from a stop, and it doesn't take off at full speed all at once: it builds up speed little by little, always at the same rate. A rock falling from a lookout doesn't fall at constant speed either: it gains the same amount of speed every second. That motion, in a straight line with constant acceleration, is UARM. How do you describe it with numbers? By the end of this page you'll know.
What UARM is
If you record a falling ball or an electric skateboard starting up, you'll see that in the first second it covers a short stretch, in the next a longer one, and in the third one even longer. How can you describe precisely a motion where the speed changes at every instant, but at a constant rate?
Check it with a simple experiment
You need a long board or ruler, some books to tilt it, a marble, and a stopwatch (the one on your cell phone works). With tape, mark distances of $10\ \text{cm}$, $40\ \text{cm}$, and $90\ \text{cm}$ from the starting point on the board. Release the marble without pushing it and measure the time it takes to reach each mark (repeat several times and take an average).
You'll see that it takes roughly $1$, $2$, and $3$ units of time (for example, $0.5\ \text{s}$, $1\ \text{s}$, and $1.5\ \text{s}$). The distances grow like $1$, $4$, $9$ while the time grows like $1$, $2$, $3$: that's the signature of a UARM. If the marble covered equal distances in equal times, it would be a uniform rectilinear motion (URM).
Explain in your own words what it means for a motion to be uniformly accelerated rectilinear motion (UARM) and give an example from everyday life.
Explain in your own words what it means for a motion to be uniformly accelerated rectilinear motion (UARM) and give an example from everyday life.
State which of these motions are approximately a UARM and justify each answer: a) A car on a motorway with the speedometer fixed at $120\ \text{km/h}$. b) A ball falling…
State which of these motions are approximately a UARM and justify each answer:
- a) A car on a motorway with the speedometer fixed at $120\ \text{km/h}$.
- b) A ball falling from a roof.
- c) A lift starting up and gaining speed uniformly.
- d) A satellite orbiting the Earth.
In equation $v = v_0 + a\times t$, state what each letter represents and the International System unit used to measure each quantity.
In equation $v = v_0 + a\times t$, state what each letter represents and the International System unit used to measure each quantity.
The velocity equation
If acceleration is the velocity gained each second, after $t$ seconds you will have gained $a \times t$. That's why all you have to do is add that to the initial velocity.
Move the initial velocity and acceleration sliders: the $v(t)$ graph is a slanted straight line (not horizontal, as in URM) and the position $x(t)$ is a curve.
A car starts from rest and accelerates at $2\ \text{m/s}^2$. Using equation $v = v_0 + a\times t$, calculate its velocity at $5\ \text{s}$.
A car starts from rest and accelerates at $2\ \text{m/s}^2$. Using equation $v = v_0 + a\times t$, calculate its velocity at $5\ \text{s}$.
A cyclist already travelling at $5\ \text{m/s}$ accelerates at $3\ \text{m/s}^2$ for $4\ \text{s}$. What speed does the cyclist reach?
A cyclist already travelling at $5\ \text{m/s}$ accelerates at $3\ \text{m/s}^2$ for $4\ \text{s}$. What speed does the cyclist reach?
An electric car goes from $8\ \text{m/s}$ to $23\ \text{m/s}$ in $5\ \text{s}$ with constant acceleration. What is its acceleration?
An electric car goes from $8\ \text{m/s}$ to $23\ \text{m/s}$ in $5\ \text{s}$ with constant acceleration. What is its acceleration?
Solving for a variable in the velocity equation
Sometimes you're not looking for $v$, but for the initial velocity, the acceleration, or the time. You solve for it as in any equation.
A car travelling at $10\ \text{m/s}$ brakes with a deceleration of $-2\ \text{m/s}^2$. How long does it take to come to a complete stop?
A car travelling at $10\ \text{m/s}$ brakes with a deceleration of $-2\ \text{m/s}^2$. How long does it take to come to a complete stop?
After $4\ \text{s}$ accelerating at $3\ \text{m/s}^2$, an electric scooter reaches $22\ \text{m/s}$. What was its initial speed when it started accelerating?
After $4\ \text{s}$ accelerating at $3\ \text{m/s}^2$, an electric scooter reaches $22\ \text{m/s}$. What was its initial speed when it started accelerating?
A train accelerates from $5\ \text{m/s}$ with an acceleration of $2.5\ \text{m/s}^2$. How long does it take to reach $20\ \text{m/s}$?
A train accelerates from $5\ \text{m/s}$ with an acceleration of $2.5\ \text{m/s}^2$. How long does it take to reach $20\ \text{m/s}$?
Free fall and vertical launch
A stone is dropped from a viewpoint. Taking $g = 10\ \text{m/s}^2$ so that the numbers are neat, what is its speed after $4\ \text{s}$ of falling?
A stone is dropped from a viewpoint. Taking $g = 10\ \text{m/s}^2$ so that the numbers are neat, what is its speed after $4\ \text{s}$ of falling?
You throw a ball vertically upwards at $30\ \text{m/s}$, as shown in the figure. With $g = 10\ \text{m/s}^2$, its speed decreases by $10\ \text{m/s}$ every second. What…
You throw a ball vertically upwards at $30\ \text{m/s}$, as shown in the figure. With $g = 10\ \text{m/s}^2$, its speed decreases by $10\ \text{m/s}$ every second.
What speed does it have $1\ \text{s}$ after launch? And after $2\ \text{s}$?
The position equation
A car starts from rest with $a = 2\ \text{m/s}^2$: in $1\ \text{s}$ it covers $1\ \text{m}$, in $2\ \text{s}$ it has covered $4\ \text{m}$, and in $3\ \text{s}$ it's already at $9\ \text{m}$. Why don't the meters grow in proportion to the time, and what formula calculates them?
A car starts from rest with an acceleration of $3\ \text{m/s}^2$ for $4\ \text{s}$. Calculate the distance it travels using $x = \frac{1}{2}a\times t^2$ and its speed at…
A car starts from rest with an acceleration of $3\ \text{m/s}^2$ for $4\ \text{s}$. Calculate the distance it travels using $x = \frac{1}{2}a\times t^2$ and its speed at the end of those $4\ \text{s}$.
A car travelling at $10\ \text{m/s}$ accelerates at $2\ \text{m/s}^2$ for $5\ \text{s}$, as shown in the figure. a) What speed does it reach? b) What distance does it…
A car travelling at $10\ \text{m/s}$ accelerates at $2\ \text{m/s}^2$ for $5\ \text{s}$, as shown in the figure.
- a) What speed does it reach?
- b) What distance does it travel in those $5\ \text{s}$? Use $x = v_0\times t + \frac{1}{2}a\times t^2$.
- c) Why is that distance not simply $10\ \text{m/s}$ multiplied by $5\ \text{s}$? Explain why.
A jet ski starts from rest and accelerates at $3\ \text{m/s}^2$ for $4\ \text{s}$; it then maintains the speed reached for another $6\ \text{s}$, as shown in the graph.…
A jet ski starts from rest and accelerates at $3\ \text{m/s}^2$ for $4\ \text{s}$; it then maintains the speed reached for another $6\ \text{s}$, as shown in the graph.
- a) What is its speed at the end of the acceleration?
- b) What distance does it travel while accelerating? Use $x = \frac{1}{2}a\times t^2$.
- c) What distance does it travel afterwards at constant speed?
- d) What is the total distance travelled?
Free fall with the position equation
In free fall we drop the object ($v_0 = 0$) and the acceleration is $g$, so the equation becomes $x = \frac{1}{2} g\,t^2$ for the distance fallen, and $v = g\,t$ for the velocity.
A flowerpot falls from a height of $45\ \text{m}$, as shown in the figure. Taking $g = 10\ \text{m/s}^2$ and knowing that in free fall the distance travelled is…
A flowerpot falls from a height of $45\ \text{m}$, as shown in the figure. Taking $g = 10\ \text{m/s}^2$ and knowing that in free fall the distance travelled is $x = \frac{1}{2}g\times t^2$:
- a) How long does it take to hit the ground?
- b) At what speed does it hit the ground?
You drop a bottle of water from a window $20\ \text{m}$ above the ground ($g = 10\ \text{m/s}^2$). Using $x = \frac{1}{2}g\times t^2$ and $v = g\times t$, calculate how…
You drop a bottle of water from a window $20\ \text{m}$ above the ground ($g = 10\ \text{m/s}^2$). Using $x = \frac{1}{2}g\times t^2$ and $v = g\times t$, calculate how long it takes to fall and at what speed it hits the ground.
From a tower, one object is dropped from a height of $5\ \text{m}$ and another from $20\ \text{m}$ ($g = 10\ \text{m/s}^2$), as shown in the figure. a) Calculate the…
From a tower, one object is dropped from a height of $5\ \text{m}$ and another from $20\ \text{m}$ ($g = 10\ \text{m/s}^2$), as shown in the figure.
- a) Calculate the time of fall for each one using $x = \frac{1}{2}g\times t^2$.
- b) Calculate the speed at which each one hits the ground.
- c) The second falls from four times the height: does it also take four times as long? Explain what actually happens.
Vertical launch upward
On the way up, gravity slows the object down: with $a = -g$, the position is $x = v_0\,t - \frac{1}{2} g\,t^2$. At the highest point $v = 0$. On the way down along the same path, it returns with the same speed it left with (if there is no air friction).
You throw a ball vertically upwards at $20\ \text{m/s}$ from the ground ($g = 10\ \text{m/s}^2$). a) How long does it take to reach maximum height, which is when its…
You throw a ball vertically upwards at $20\ \text{m/s}$ from the ground ($g = 10\ \text{m/s}^2$).
- a) How long does it take to reach maximum height, which is when its velocity becomes zero?
- b) Using $x = v_0\times t - \frac{1}{2}g\times t^2$, calculate that maximum height.
A toy rocket ascends with an initial velocity of $30\ \text{m/s}$ and, once the engine turns off, only gravity acts ($g = 10\ \text{m/s}^2$, ignoring friction). a) How…
A toy rocket ascends with an initial velocity of $30\ \text{m/s}$ and, once the engine turns off, only gravity acts ($g = 10\ \text{m/s}^2$, ignoring friction).
- a) How long does it take to reach maximum height?
- b) What maximum height does it reach? Use $x = v_0\times t - \frac{1}{2}g\times t^2$.
- c) How long does it take in total to return to the starting point?
- d) At what speed does it return? Comment on the result by comparing it with the launch speed.
You want a ball to hit the ground with a speed of $40\ \text{m/s}$ by dropping it from a certain height ($g = 10\ \text{m/s}^2$, ignoring friction). a) How long must the…
You want a ball to hit the ground with a speed of $40\ \text{m/s}$ by dropping it from a certain height ($g = 10\ \text{m/s}^2$, ignoring friction).
- a) How long must the fall take?
- b) From what height must you drop it? Use $x = \frac{1}{2}g\times t^2$.
- c) Check your results by calculating the final speed from the data you obtained.
When the time comes from a quadratic equation
If $v_0 \neq 0$ and you're asked for the time from a distance, the time ends up in both $t$ and $t^2$: it's a quadratic equation. You move everything to one side, simplify, and solve by factoring or with the quadratic formula.
From a balcony $10\ \text{m}$ above the ground, you throw a ball vertically downwards at $5\ \text{m/s}$ ($g = 10\ \text{m/s}^2$). The distance travelled satisfies…
From a balcony $10\ \text{m}$ above the ground, you throw a ball vertically downwards at $5\ \text{m/s}$ ($g = 10\ \text{m/s}^2$). The distance travelled satisfies $10 = 5\times t + 5\times t^2$.
- a) Solve the equation and calculate how long it takes to hit the ground.
- b) At what speed does it hit the ground?
- c) If instead you threw it upwards at $5\ \text{m/s}$, the equation would be $10 = -5\times t + 5\times t^2$. Solve it and compare the two times: what is the difference and what causes it?
From a bridge $20\ \text{m}$ above the water, you throw a stone vertically upwards at $15\ \text{m/s}$ ($g = 10\ \text{m/s}^2$), as shown in the figure. Taking the water…
From a bridge $20\ \text{m}$ above the water, you throw a stone vertically upwards at $15\ \text{m/s}$ ($g = 10\ \text{m/s}^2$), as shown in the figure. Taking the water as the origin, the height of the stone satisfies $y = 20 + 15\times t - 5\times t^2$.
- a) At what height is it at time $t = 1\ \text{s}$?
- b) What speed does it have at $t = 2\ \text{s}$? Interpret the sign of the result.
- c) Solve $20 + 15\times t - 5\times t^2 = 0$ to find the time at which it falls into the water.
- d) At what speed does it hit the water?
Braking and road safety
To brake, the acceleration is negative and the car stops when $v = 0$. First you calculate the time with $v = v_0 + a\,t$ and then the distance with $x = v_0 t + \frac{1}{2} a t^2$.
Go back to the beginning: the velocity of the bus or the rock is described by $v = v_0 + a\,t$, and the distance they cover by $x = x_0 + v_0 t + \frac{1}{2} a t^2$.
A car travelling at $20\ \text{m/s}$ brakes with an acceleration of $-4\ \text{m/s}^2$, as shown in the figure. a) How long does it take to stop? b) What distance does…
A car travelling at $20\ \text{m/s}$ brakes with an acceleration of $-4\ \text{m/s}^2$, as shown in the figure.
- a) How long does it take to stop?
- b) What distance does it travel while braking? Use $x = v_0\times t + \frac{1}{2}a\times t^2$.
- c) Another car travels at twice the speed, $40\ \text{m/s}$, and brakes with the same acceleration. Calculate its braking distance. Is it double that of the first car? Comment on the result and its importance for road safety.
A driver's reaction time is $0.5\ \text{s}$. She is travelling at $20\ \text{m/s}$ and, on seeing an obstacle, brakes with an acceleration of $-5\ \text{m/s}^2$, as…
A driver's reaction time is $0.5\ \text{s}$. She is travelling at $20\ \text{m/s}$ and, on seeing an obstacle, brakes with an acceleration of $-5\ \text{m/s}^2$, as shown in the figure.
- a) What distance does she travel during her reaction time, before applying the brakes?
- b) Once she brakes, how long does it take to stop?
- c) What distance does she travel while braking? Use $x = v_0\times t + \frac{1}{2}a\times t^2$.
- d) What is the total stopping distance? What conclusion do you draw regarding driving?