Exercises: Uniformly accelerated rectilinear motion

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.

A car starts from rest and accelerates at $2\ \text{m/s}^2$. Using equation $v = v_0 + a\cdot t$, 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 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?

State which of these motions are approximately a UARM and justify each answer:

1. a) A car on a motorway with the speedometer fixed at $120\ \text{km/h}$.
2. b) A ball falling from a roof.
3. c) A lift starting up and gaining speed uniformly.
4. d) A satellite orbiting the Earth.

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 $2\ \text{s}$?

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?

In equation $v = v_0 + a\cdot t$, state what each letter represents and the International System unit used to measure each quantity.

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?

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 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 $x = \frac{1}{2}g\cdot t^2$:

1. a) How long does it take to hit the ground?
2. b) At what speed does it hit the ground?

You throw a ball vertically upwards at $20\ \text{m/s}$ from the ground ($g = 10\ \text{m/s}^2$).

1. a) How long does it take to reach maximum height, which is when its velocity becomes zero?
2. b) Using $x = v_0\cdot t - \frac{1}{2}g\cdot t^2$, calculate that maximum height.

You drop a bottle of water from a window $20\ \text{m}$ above the ground ($g = 10\ \text{m/s}^2$). Using $x = \frac{1}{2}g\cdot t^2$ and $v = g\cdot t$, calculate how long it takes to fall and at what speed it hits the ground.

A car starts from rest with an acceleration of $3\ \text{m/s}^2$ for $4\ \text{s}$. Calculate the distance it travels using $x = \frac{1}{2}a\cdot t^2$ and its speed at the end of those $4\ \text{s}$.

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.

1. a) Calculate the time of fall for each one using $x = \frac{1}{2}g\cdot t^2$.
2. b) Calculate the speed at which each one hits the ground.
3. c) The second falls from four times the height: does it also take four times as long? Explain what actually happens.

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).

1. a) How long does it take to reach maximum height?
2. b) What maximum height does it reach? Use $x = v_0\cdot t - \frac{1}{2}g\cdot t^2$.
3. c) How long does it take in total to return to the starting point?
4. d) At what speed does it return? Comment on the result by comparing it with the launch speed.

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 $10 = 5\cdot t + 5\cdot t^2$.

1. a) Solve the equation and calculate how long it takes to hit the ground.
2. b) At what speed does it hit the ground?
3. c) If instead you threw it upwards at $5\ \text{m/s}$, the equation would be $10 = -5\cdot t + 5\cdot t^2$. Solve it and compare the two times: what is the difference and what causes 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.

1. a) What speed does it reach?
2. b) What distance does it travel in those $5\ \text{s}$? Use $x = v_0\cdot t + \frac{1}{2}a\cdot t^2$.
3. c) Why is that distance not simply $10\ \text{m/s}$ multiplied by $5\ \text{s}$? Explain why.

A car travelling at $20\ \text{m/s}$ brakes with an acceleration of $-4\ \text{m/s}^2$, as shown in the figure.

1. a) How long does it take to stop?
2. b) What distance does it travel while braking? Use $x = v_0\cdot t + \frac{1}{2}a\cdot t^2$.
3. 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 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.

1. a) What is its speed at the end of the acceleration?
2. b) What distance does it travel while accelerating? Use $x = \frac{1}{2}a\cdot t^2$.
3. c) What distance does it travel afterwards at constant speed?
4. d) What is the total distance travelled?

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).

1. a) How long must the fall take?
2. b) From what height must you drop it? Use $x = \frac{1}{2}g\cdot t^2$.
3. c) Check your results by calculating the final speed from the data you obtained.

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 $y = 20 + 15\cdot t - 5\cdot t^2$.

1. a) At what height is it at time $t = 1\ \text{s}$?
2. b) What speed does it have at $t = 2\ \text{s}$? Interpret the sign of the result.
3. c) Solve $20 + 15\cdot t - 5\cdot t^2 = 0$ to find the time at which it falls into the water.
4. d) At what speed does it hit the water?

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.

1. a) What distance does she travel during her reaction time, before applying the brakes?
2. b) Once she brakes, how long does it take to stop?
3. c) What distance does she travel while braking? Use $x = v_0\cdot t + \frac{1}{2}a\cdot t^2$.
4. d) What is the total stopping distance? What conclusion do you draw regarding driving?