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Motion and Forces

Overview

This topic develops your understanding of how and why objects move. It connects the description of motion (kinematics) with the forces that cause motion, using Newton’s Laws as a foundation for analyzing dynamics in real-world and theoretical situations.

Contents

  • Kinematics and uniformly accelerated motion
  • Interpreting motion graphs
  • Mass and inertia
  • Linear momentum
  • Newton’s Laws of Motion

Learning Outcomes

By the end of this section, students should be able to:

  • Define and use terms: displacement, speed, velocity, and acceleration
  • Use and interpret motion-time and velocity-time graphs
  • Derive and apply the kinematic equations for uniformly accelerated motion
  • Understand mass as a measure of inertia
  • Define and use momentum in practical situations
  • Apply Newton’s Laws to explain and predict motion
  • Solve problems using Newton’s Second Law: F = m × a

1. Kinematics and Uniformly Accelerated Motion

Key concepts:

  • Displacement (s): change in position (vector; includes direction).
  • Speed (v): scalar rate of motion; speed = distance / time.
  • Velocity (v): vector rate of change of position (magnitude + direction).
  • Acceleration (a): rate of change of velocity; a = Δv / Δt.

Equations of motion (constant acceleration):

v = u + a × t        s = u × t + 0.5 × a × t2        v2 = u2 + 2 × a × s

Symbols: u = initial velocity, v = final velocity, a = acceleration, s = displacement, t = time.

Vector relation v = u + at
Vector relation: v = u + a × t
Displacement as area under a velocity-time graph
Displacement from v–t graph: s = u × t + 0.5 × a × t2
Illustration of v^2 = u^2 + 2as
When t is unknown: v2 = u2 + 2 × a × s
Worked Example: A car accelerates from 10 m × s-1 to 30 m × s-1 in 5 s. Find its acceleration.
Use a = (v – u) / t = (30 – 10) / 5 = 4.0 m × s-2.
Learning-Check 1: A ball is thrown upward at 15 m × s-1. How long until it reaches its highest point?
Answer: At peak height, v = 0. Taking upward as positive, a = -g = -9.81 m × s-2.
t = (v – u) / a = (0 – 15) / -9.81 ≈ 1.53 s.
Learning-Check 2: A train accelerates uniformly from rest to 25 m × s-1 in 10 s. Find the displacement in this time.
Answer: Acceleration a = (25 – 0) / 10 = 2.5 m × s-2.
Displacement s = u × t + 0.5 × a × t2 = 0 + 0.5 × 2.5 × 100 = 125 m.

Exam tip: Choose a clear sign convention (e.g., upward positive) and stick to it; check units at each step (m, s, m × s-1, m × s-2).

2. Interpreting Motion Graphs

Displacement–time graph (s–t):

  • Slope = instantaneous velocity: v = Δs / Δt for a straight segment, v(t) = ds / dt in general.
  • Horizontal line ⇒ v = 0 (object at rest). Steeper line ⇒ larger speed.
  • Positive slope ⇒ motion in + direction; negative slope ⇒ motion in − direction.
Displacement-time graph
Displacement–time: slope gives velocity.

Velocity–time graph (v–t):

  • Slope = acceleration: a = Δv / Δt for a straight segment, a(t) = dv / dt in general.
  • Area under the curve = displacement over an interval: s = ∫ v(t) dt.
  • Horizontal line ⇒ constant velocity (a = 0). A rising line through v = 0 ⇒ speeding up in the + direction.
Velocity-time graph
Velocity–time: slope gives acceleration; area gives displacement.
Worked Example: A v–t graph is a straight line from v = 0 at t = 0 to v = 20 m × s-1 at t = 4 s.
  • Acceleration (slope): a = Δv / Δt = (20 – 0) / (4 – 0) = 5.0 m × s-2.
  • Displacement (area of triangle): s = 0.5 × base × height = 0.5 × 4 × 20 = 40 m.
Learning-Check 1: On a v–t graph, what does a horizontal line represent?
Answer: Constant velocity (a = 0).
Learning-Check 2: What physical quantity is represented by the area under a v–t graph?
Answer: Displacement over that time interval (s = ∫ v(t) dt).

3. Mass and Inertia

Mass (m) quantifies an object’s resistance to changes in its motion (inertia). For a given net force, acceleration is smaller when mass is larger: a = ∑ F / m.

Inertia is the natural tendency of an object to maintain its current state of motion (rest or constant velocity). More mass ⇒ more inertia ⇒ harder to start, stop, or turn.

Units & types of mass: SI unit is kilogram (kg). In everyday mechanics we treat inertial mass (resistance to acceleration) and gravitational mass (weight generation W = m × g) as numerically identical.

Mass vs Weight: Mass (kg) is intrinsic; weight is a gravitational force: W = m × g (Newtons). On the Moon, local g is smaller, so weight decreases while mass stays constant.

Worked Example: Two carts are pushed by the same horizontal force F = 12 N. Cart A has mA = 3.0 kg; Cart B has mB = 6.0 kg.
  • Cart A: aA = F / mA = 12 / 3.0 = 4.0 m × s-2.
  • Cart B: aB = F / mB = 12 / 6.0 = 2.0 m × s-2.
  • Conclusion: The lighter cart accelerates twice as fast under the same force (less inertia).
Learning-Check 1: Two carts, one light and one heavy, are pushed with equal force. Which accelerates more, and why?
Answer: The lighter cart. From a = F / m, smaller mass yields a larger acceleration due to lower inertia.
Learning-Check 2: What physical property resists changes in velocity? Give its SI unit and a common equation where it appears.
Answer: Inertia (quantified by mass, unit kg). It appears in ∑ F = m × a and W = m × g.

4. Linear Momentum

Definition (vector): Linear momentum p is defined as p = m × v. SI Unit: kg·m·s-1.

Impulse–momentum theorem: The impulse J delivered by a net force equals the total change in linear momentum:

J = ∫ Fnet dt = Δp = pfinal – pinitial

For a constant average force over time Δt: Favg = Δp / Δt.

Newton’s Second Law (momentum form): Fnet = dp / dt. For constant mass, this reduces to Fnet = m × a.

Conservation of Momentum: In an isolated system (where net external impulse = 0), total system momentum is strictly conserved: ∑ pinitial = ∑ pfinal.

Impulse changes momentum graph
Impulse changes momentum: area under F–t curve equals Δp.
Worked Example 1 (Wall Impact): A 2.0 kg trolley moves at +3.0 m × s-1, hits a wall, and stops in 0.10 s.
  • Initial momentum: pi = 2.0 × (+3.0) = +6.0 kg·m·s-1.
  • Final momentum: pf = 0 ⇒ Δp = 0 – 6.0 = -6.0 kg·m·s-1.
  • Average force on trolley: Favg = Δp / Δt = -6.0 / 0.10 = -60 N (directed opposite incoming motion).
Worked Example 2 (Inelastic Collision): A 0.40 kg cart moving at +2.5 m × s-1 sticks to a 0.60 kg cart at rest. Find common final speed.
  • Initial momentum: pi = (0.40 × 2.5) + (0.60 × 0) = +1.00 kg·m·s-1.
  • Final combined mass = 1.00 kg ⇒ pf = 1.00 × vf.
  • Conservation: pf = pi ⇒ vf = 1.00 / 1.00 = 1.00 m × s-1.
Learning-Check 1: A 0.50 kg ball rolls at 8.0 m × s-1. What is its momentum?
Answer: p = m × v = 0.50 × 8.0 = 4.0 kg·m·s-1.
Learning-Check 2: A constant force of 20 N acts for 0.20 s on a 0.40 kg puck initially at rest. What impulse is delivered and what final speed is reached?
Answer: Impulse J = F × Δt = 20 × 0.20 = 4.0 N·s.
Final speed v = Δp / m = 4.0 / 0.40 = 10 m × s-1.

5. Newton’s Laws of Motion

  • Newton’s First Law: An object continues in its state of rest or uniform motion in a straight line unless acted upon by a net external force.
  • Newton’s Second Law: The rate of change of momentum of a body is directly proportional to the applied net force and takes place in the direction of that force (∑ F = m × a).
  • Newton’s Third Law: When two bodies interact, the forces they exert on each other are equal in magnitude and opposite in direction (action-reaction pairs).

Key Concepts Recap

Core relations (kinematics, Newton’s laws, and momentum)
ConceptFormulaWhen / Notes
Displacement (s)s = u × t + 0.5 × a × t2Uniform acceleration; area under v–t graph.
Final velocity (v)v = u + a × tVector relation v = u + a × t.
Time-free kinematicv2 = u2 + 2 × a × sUse when time t is unknown.
Average accelerationaavg = Δv / ΔtSlope of v–t graph on straight segments.
Displacement from v–ts = ∫ v(t) dtArea under v–t curve; signed direction matters.
Newton’s 2nd Law∑ F = m × aMomentum form: ∑ F = dp / dt.
WeightW = m × gNear Earth, g ≈ 9.81 m × s-2.
Linear momentump = m × vUnit: kg·m·s-1; vector quantity.
ImpulseJ = ∫ F dt = ΔpFor constant force: J = F × Δt.

Interactive Simulator: Newton’s 2nd Law & Stopping Distance

Adjust object mass, initial velocity, and applied braking force to calculate deceleration, stopping time, and total displacement in real-time.

Inertial mass of vehicle
Initial speed before braking
Retarding friction force
Calculation Formulas
a = F / m  |  tstop = u / a  |  sstop = u2 / (2 × a)
Deceleration Rate
5.0 m/s²
Stopping Time
5.0 s
Stopping Distance
62.5 m

Interactive Review Questions

Click to Reveal: Why does doubling initial speed quadruple the required braking distance under constant force?
Explanation: From the kinematic relation v2 = u2 + 2 × a × s, setting final velocity v = 0 yields s = u2 / (2 × a). Stopping distance scales with the square of initial speed (u2).
Click to Reveal: How does Newton’s Third Law apply when a heavy truck collides with a compact car?
Explanation: By Newton’s Third Law, the force exerted by the truck on the car is identical in magnitude to the force exerted by the car on the truck. However, because the car has far less mass, it experiences a significantly higher acceleration (a = F / m).

Frequently Asked Questions

What is the difference between average velocity and instantaneous velocity?

Average velocity is total displacement divided by total time over a finite interval (Δs / Δt). Instantaneous velocity is the exact derivative of position with respect to time (ds / dt) at a specific instant.

How do you distinguish between elastic and inelastic collisions?

Total momentum is conserved in both types of collisions. In elastic collisions, total mechanical kinetic energy is also conserved. In inelastic collisions, kinetic energy is partially converted into thermal, acoustic, or deformation energy.

Why is mass considered an intrinsic property while weight is variable?

Mass measures the intrinsic quantity of matter and resistance to acceleration (inertia), remaining constant anywhere in the universe. Weight is the local gravitational force (W = m × g), which varies depending on local gravitational field strength.

Additional Resources

To deepen your understanding, you are strongly encouraged to explore the materials provided on the following pages:

End-of-Page Comprehensive Practice Module

Section 1: Foundational Core Concepts

  1. What is the primary difference between distance and displacement?
    Answer: Distance is the scalar total path length traveled. Displacement is the vector straight-line distance and direction from the initial to final position.
  2. Define average speed versus instantaneous velocity.
    Answer: Average speed is total distance divided by total time. Instantaneous velocity is the vector time-derivative of position (ds / dt) at a specific instant.
  3. What does the slope of a displacement–time graph represent?
    Answer: The instantaneous velocity of the object.
  4. What physical quantity is given by the area under a velocity–time graph?
    Answer: The total displacement over that time interval.
  5. State the three constant-acceleration kinematics equations.
    Answer: v = u + a × t,   s = u × t + 0.5 × a × t2,   v2 = u2 + 2 × a × s.
  6. Define linear momentum and state its SI units.
    Answer: Linear momentum is the product of mass and velocity (p = m × v), measured in kg·m·s-1 (or N·s).
  7. State Newton’s First Law of Motion.
    Answer: An object remains at rest or continues in motion at a constant velocity unless acted upon by a non-zero net external force.
  8. State Newton’s Second Law of Motion in terms of momentum.
    Answer: Net external force equals the rate of change of linear momentum (∑ F = dp / dt).

Section 2: Analytical & Scenario-Based Problems

  1. A runner completes one full lap around a 400 m circular track in 50 seconds. Compare their average speed and average velocity.
    Answer: Average speed = 400 m / 50 s = 8.0 m/s. Because the runner returns to the start line, total displacement = 0 m, so average velocity = 0.0 m/s.
  2. On a velocity–time graph, what does a straight line with a negative slope indicate?
    Answer: Uniform negative acceleration (deceleration if moving in the positive direction, or speeding up if moving in the negative direction).
  3. Why does a heavy freight train require a much greater stopping distance than a passenger car moving at the same speed?
    Answer: The freight train possesses significantly greater mass and inertia, resulting in a much larger initial momentum and kinetic energy that requires far more braking work to bring to rest.
  4. Explain how safety airbags in cars reduce impact forces during a crash using impulse concepts.
    Answer: Airbags increase the impact time duration (Δt) during which momentum changes (Δp). Since impulse J = Favg × Δt = Δp, increasing Δt reduces average force Favg.
  5. In an Atwood machine with two unequal masses m1 > m2, why is string tension less than m1 × g but greater than m2 × g?
    Answer: For m1 to accelerate downward, m1 × g – T = m1 × a ⇒ T < m1 × g. For m2 to accelerate upward, T – m2 × g = m2 × a ⇒ T > m2 × g.
  6. Explain why a person feels heavier when an elevator accelerates upward.
    Answer: The scale reads normal force N. To accelerate upward, net force N – m × g = m × a ⇒ N = m × (g + a), which exceeds true weight m × g.
  7. If no external forces act on an exploding firework shell in mid-air, what happens to the motion of its center of mass?
    Answer: Internal explosive forces cancel in pairs by Newton’s Third Law. The center of mass continues along its original parabolic path completely unaffected.
  8. Why is kinetic energy lost in a perfectly inelastic collision even though linear momentum is conserved?
    Answer: Momentum conservation requires total velocity to drop as combined mass increases. The lost kinetic energy is converted into internal thermal energy, sound, and material deformation work.

Section 3: Numerical Problems with Step-by-Step Solutions

  1. A car accelerates uniformly from 10.0 m/s to 30.0 m/s in 5.0 seconds. Calculate its acceleration and total displacement.
    Solution:
    1. Acceleration: a = (v – u) / t = (30.0 – 10.0) / 5.0 = 4.0 m/s2
    2. Displacement: s = u × t + 0.5 × a × t2 = (10.0 × 5.0) + (0.5 × 4.0 × 25.0) = 50.0 + 50.0 = 100.0 m
  2. A stone is dropped from rest off a 20.0 m high bridge. Calculate the time taken to hit the water (take g = 9.81 m/s2).
    Solution:
    Use s = u × t + 0.5 × g × t2 with u = 0
    20.0 = 0.5 × 9.81 × t2  ⇒  4.905 × t2 = 20.0  ⇒  t2 = 4.077  ⇒  t = 2.02 s
  3. A car moving at 25.0 m/s applies brakes with constant deceleration a = -5.0 m/s2. Calculate time to stop and braking distance.
    Solution:
    1. Stopping time: v = u + a × t ⇒ 0 = 25.0 – 5.0 × t ⇒ t = 5.0 s
    2. Stopping distance: v2 = u2 + 2 × a × s ⇒ 0 = 625.0 – 10.0 × s ⇒ s = 62.5 m
  4. A 1000 kg car traveling at 20.0 m/s impacts a barrier and stops in 0.10 s. Calculate the average retarding force.
    Solution:
    1. Momentum change: Δp = m × (v – u) = 1000 × (0 – 20.0) = -20,000 kg·m/s
    2. Average force: Favg = Δp / Δt = -20,000 / 0.10 = -200,000 N (-200 kN)
  5. A 0.010 kg bullet is fired from a 5.0 kg rifle at 400.0 m/s. Calculate the recoil speed of the rifle.
    Solution:
    Conservation of momentum: mrifle × vrecoil + mbullet × vbullet = 0
    5.0 × vrecoil + (0.010 × 400.0) = 0 ⇒ 5.0 × vrecoil = -4.0
    vrecoil = -4.0 / 5.0 = -0.80 m/s (0.80 m/s backward)
  6. An Atwood machine has masses m1 = 5.0 kg and m2 = 3.0 kg over a frictionless pulley. Calculate acceleration and string tension (g = 9.81 m/s2).
    Solution:
    1. Acceleration: a = ((m1 – m2) × g) / (m1 + m2) = ((5.0 – 3.0) × 9.81) / (5.0 + 3.0) = 19.62 / 8.0 = 2.45 m/s2
    2. Tension: T = m1 × (g – a) = 5.0 × (9.81 – 2.45) = 5.0 × 7.36 = 36.8 N
  7. A 4.0 kg block slides down a 30.0° incline with kinetic friction coefficient μk = 0.20. Calculate acceleration (g = 9.81 m/s2).
    Solution:
    1. Downhill gravity force: Fg = m × g × sin(30°) = 4.0 × 9.81 × 0.50 = 19.62 N
    2. Friction force: fk = μk × m × g × cos(30°) = 0.20 × 4.0 × 9.81 × 0.8660 = 6.79 N
    3. Net force: Fnet = 19.62 – 6.79 = 12.83 N
    4. Acceleration: a = Fnet / m = 12.83 / 4.0 = 3.21 m/s2
  8. A 600 kg elevator accelerates upward at 1.50 m/s2. Calculate cable tension (g = 9.81 m/s2).
    Solution:
    Vertical force balance: T – m × g = m × a ⇒ T = m × (g + a)
    T = 600 × (9.81 + 1.50) = 600 × 11.31 = 6786 N (6.79 kN)
Last updated: 05 Aug 2026