Motion with Constant Acceleration

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:

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

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 traveling at $5\ \text{m/s}$ accelerates at $3\ \text{m/s}^2$ for $4\ \text{s}$. What speed does the cyclist reach?

A car traveling 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 $v = v_0 + a\times t$, state what each letter represents and the International System of Units (SI) 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 stone is dropped from a lookout point. 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 upward 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 $2\ \text{s}$?

You throw a ball vertically upward 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 $x = v_0\times t - \frac{1}{2}g\times t^2$, calculate that maximum height.

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 $x = \frac{1}{2}g\times t^2$.
3. c) Check your results by calculating the final speed from the data you obtained.

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.

1. a) Calculate the time of fall for each one using $x = \frac{1}{2}g\times t^2$.
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 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 traveling 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 $x = v_0\times t + \frac{1}{2}a\times t^2$.
3. c) Why is that distance not simply $10\ \text{m/s}$ multiplied by $5\ \text{s}$? Explain why.

A car traveling 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 $x = v_0\times t + \frac{1}{2}a\times t^2$.
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 $x = \frac{1}{2}a\times t^2$.
3. c) What distance does it travel afterwards at constant speed?
4. d) What is the total distance traveled?

A driver's reaction time is $0.5\ \text{s}$. She is traveling 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 $x = v_0\times t + \frac{1}{2}a\times t^2$.
4. d) What is the total stopping distance? What conclusion do you draw regarding driving?