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Physical Science B — Forces, Energy and Waves (California)

Curriculum

  • 4 Sections
  • 20 Lessons
  • Lifetime
Expand all sectionsCollapse all sections
  • Unit 1: Motion, Forces and Collision Safety
    5
    • 1.1
      Crash Cushions at the Off-Ramp
      50 mins
    • 1.2
      Newton’s Second Law with Carts
      100 mins
    • 1.3
      Momentum Before and After a Collision
      50 mins
    • 1.4
      Impulse: Stretching the Stop
      100 mins
    • 1.5
      Performance Task — Crash Cushion Design Challenge
      200 mins
  • Unit 2: Gravity, Electricity and Magnetism
    5
    • 2.1
      Satellites Over Vandenberg: Gravity at a Distance
      50 mins
    • 2.2
      Electromagnets and Moving Magnets
      100 mins
    • 2.3
      Coulomb’s Law: Electric Force Versus Gravity
      50 mins
    • 2.4
      Energy in Fields: Motors and Generators
      50 mins
    • 2.5
      Performance Task — From Wind to Wire
      150 mins
  • Unit 3: Energy Conservation and Transformation
    5
    • 3.1
      Storing Sunshine by Pumping Water Uphill
      50 mins
    • 3.2
      Mixing Hot and Cold Water
      100 mins
    • 3.3
      Where Did the Energy Go? Modeling Bouncing Balls
      50 mins
    • 3.4
      Designing a Solar Oven
      100 mins
    • 3.5
      Performance Task — Energy Device Design Report
      200 mins
  • Unit 4: Waves and Information
    5
    • 4.1
      Warning Before the Shaking
      50 mins
    • 4.2
      Waves on a Spring
      100 mins
    • 4.3
      Wave or Particle? Light, Photons and Health Claims
      100 mins
    • 4.4
      Digital Signals and the Devices That Carry Them
      50 mins
    • 4.5
      Performance Task — ShakeAlert Technical Explainer
      150 mins

Crash Cushions at the Off-Ramp

Unit 1  ·  Phenomenon Launch & Questioning  ·  Lesson 1 of 20

Crash Cushions at the Off-Ramp

HS-PS2-1PHYSSCIB-CA
By the end of this lesson I can…

describe motion with position, velocity and acceleration, calculate acceleration from a change in velocity, and explain why an object’s motion changes only when a net force acts on it.

Instruction

The phenomenon. Drive any California freeway and look where an off-ramp splits from the main lanes. In front of the concrete divider there is often a row of yellow plastic barrels filled with sand, or a long folding metal-and-plastic cushion. Caltrans calls these impact attenuators. A car that hits the bare concrete stops almost instantly; a car that hits the barrels stops over a longer distance and time. Why does that difference matter so much to the people inside? To answer, we need a language for motion and a rule connecting motion to force.

Describing motion. Position is where an object is, measured from a reference point. Velocity is how fast position changes, with a direction: average velocity = change in position ÷ time. A car that travels 150 m north in 6.0 s has an average velocity of 25 m/s north. (25 m/s is 90 km/h, about 56 mph.) Acceleration is how fast velocity changes: acceleration = change in velocity ÷ time. Its unit is meters per second per second, m/s2.

Worked example. A car speeds up from 0 to 26.8 m/s (60 mph) in 8.0 s. Its acceleration is (26.8 − 0) ÷ 8.0 = 3.35 m/s2: every second it gains 3.35 m/s. Slowing down is acceleration too, in the direction opposite the motion. A car moving at 25 m/s that stops in 0.10 s against concrete has an acceleration of (0 − 25) ÷ 0.10 = −250 m/s2, about 25 times the acceleration of gravity (9.8 m/s2). The same car stopped by a cushion over 0.50 s has an acceleration of −50 m/s2, five times smaller. These numbers are illustrative, but the comparison is the point.

Graphs. On a velocity-time graph, a horizontal line means constant velocity (zero acceleration), a line sloping up means speeding up, and the slope of the line is the acceleration. A steep drop to zero is a sudden stop; a gentle slope to zero is a gradual one. You will read and draw these graphs all unit.

Forces change motion. A force is a push or a pull, measured in newtons (N). Several forces usually act on an object at once: gravity, the push of the road (normal force), friction, air resistance. What matters is the net force, the combined effect with directions taken into account. Newton’s first law says that an object keeps moving at constant velocity, or stays at rest, unless a net force acts on it. This tendency is called inertia. A passenger in a crash keeps moving forward at 25 m/s until something (a seatbelt, an airbag, the dashboard) exerts a force to stop them.

Newton’s second law. The acceleration of an object is proportional to the net force and inversely proportional to its mass: a = Fnet ÷ m, or Fnet = ma. For a 70 kg person to decelerate at 250 m/s2, the net force on them must be 70 × 250 = 17,500 N, roughly 25 times their weight. At 50 m/s2 it is 3,500 N. So a cushion that reduces the acceleration also reduces the force on every person in the car. That is the logic of every safety device in this unit.

Your questions. Build the question board. Which questions could you test with carts and a ramp? Which need published crash-test data? Which need a model, such as momentum, to answer?

Vocabulary in context

  • velocity — The rate of change of position, including direction; change in position divided by time.
  • acceleration — The rate of change of velocity; change in velocity divided by time, in m/s2.
  • net force — The overall force on an object when all the forces acting on it are combined, taking direction into account.
  • inertia — The tendency of an object to keep its state of motion unless a net force acts on it.
  • impact attenuator — A roadside device that absorbs a vehicle’s energy and lengthens its stopping time in a crash.

Formative check

Work through these before moving on. They are not graded — they tell you, and your teacher, whether the standard below has landed yet.

Make a prediction

Two identical cars moving at 25 m/s stop, one against concrete in 0.10 s and one in a crash cushion in 0.50 s. How does the force on each driver compare?

About five times smaller. The change in velocity is the same, but it happens over five times as long, so the acceleration, and by <em>F</em> = <em>ma</em> the force, is five times smaller.
Fill in the blank

A bike goes from 2 m/s to 8 m/s in 3 s, so its acceleration is m/s2.

+50 XP

A hockey puck slides across smooth ice at constant velocity. What is the net force on it?

Constant velocity means zero acceleration, so the net force is zero (Newton’s first law).

Simulation & tools

Net force and acceleration

Open the Acceleration screen and turn on Forces, Values, Mass and Acceleration. Set friction to zero. (1) Apply 100 N to the 50 kg crate and record the acceleration. (2) Double the applied force and record it again. (3) Stack a second object on the crate, record the total mass, apply 200 N and record the acceleration. (4) Write the pattern you see as a sentence and as an equation. No device? Use a = F ÷ m for 100 N and 200 N on 50 kg and 200 N on 100 kg and answer the same question.

Open in a new tab ↗  ·  PhET Interactive Simulations, University of Colorado Boulder · CC BY 4.0

Quick self-check

How confident are you that you can describe motion, calculate acceleration, and explain changes in motion with net force?

Not yetVery confident

Practice

Work these on paper or in your notebook, then open Check your answer. Aim for all of Fluency and Application; try at least one Challenge.

Printable version: this unit’s practice workbook (PDF)

Fluency

Build speed and accuracy with the core skill.

  1. A car moving at 5 m/s speeds up at 2 m/s2 for 6 s. What is its final speed?
    Check your answer
    Answer: 17.0 m/s
    v = v0 + at.
  2. A car starting from rest speeds up at 3 m/s2 for 9 s. What is its final speed?
    Check your answer
    Answer: 27.0 m/s
    v = v0 + at.
  3. A car moving at 5 m/s speeds up at 4 m/s2 for 3 s. What is its final speed?
    Check your answer
    Answer: 17.0 m/s
    v = v0 + at.
  4. Convert 60 °C to kelvin.
    Check your answer
    Answer: 333.15 K
    K = °C + 273.15.
  5. A runner covers 120 m in 8.0 s. What is her average speed?
    Check your answer
    Answer: 15 m/s
    120 ÷ 8.0 = 15.
  6. A skateboarder speeds up from 0 to 8 m/s in 4 s. What is the acceleration?
    Check your answer
    Answer: 2 m/s2
    (8 − 0) ÷ 4 = 2.
  7. On a velocity-time graph, what does a horizontal line mean? A line sloping down to zero?
    Check your answer
    Answer: Constant velocity (zero acceleration); slowing down to a stop.

Application

Use the skill in context. Show your reasoning.

  1. A car moving at 25 m/s stops against a crash cushion in 0.50 s. Find its acceleration and the net force on a 70 kg passenger during the stop.
    Check your answer
    Answer: −50 m/s2; 3,500 N (opposite the motion)
    (0 − 25) ÷ 0.50 = −50 m/s2. F = 70 × 50 = 3,500 N.
  2. Explain with Newton’s first law why an unbelted passenger hits the dashboard when a car stops suddenly.
    Check your answer
    Answer: The passenger keeps moving forward at the car’s original speed (inertia) because no force acts on them to slow them until they hit the dashboard. The seatbelt supplies that force earlier and over a longer time.
  3. A 60 kg cyclist and her 12 kg bike accelerate at 1.5 m/s2. What net force acts on the cyclist-and-bike system? If friction and air resistance total 20 N, what forward force must the road exert on the tires?
    Check your answer
    Answer: 108 N net; 128 N forward
    Fnet = 72 × 1.5 = 108 N. Forward force = 108 + 20 = 128 N.

Challenge

Stretch problems. Expect to think before you write.

  1. Sketch the velocity-time graphs for the car hitting concrete (stops in 0.10 s) and the car hitting the cushion (stops in 0.50 s), both from 25 m/s. What feature of each graph shows the acceleration, and what does the area under each graph represent?
    Check your answer
    Answer: Both are straight lines falling from 25 m/s to 0; the slope is the acceleration (−250 and −50 m/s2). The area under each line is the stopping distance: 1.25 m and 6.25 m.
    Area of a triangle: ½ × 25 × 0.10 = 1.25 m; ½ × 25 × 0.50 = 6.25 m.

Review

Keep earlier skills sharp.

  1. How many moles are in 100.0 g of O2?
    Check your answer
    Answer: 3.125 mol
    Divide by the molar mass, 32.00 g/mol.
  2. A wood sample has 25% of the carbon-14 of living wood. Using a half-life of 5,730 years, estimate its age.
    Check your answer
    Answer: ≈ 11,460 years
    t = 5730 × log2(100/percent remaining).

CA NGSS and CCSS literacy standards addressed: HS-PS2-1, SEP.1, CCC.2, RST.9-10.4

UC A-G Area D pillar: Describing motion and connecting changes in motion to net force

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