Motion in a Straight Line · Lesson 10 of 10

Chapter Summary and Practice

“Connect the complete chapter, diagnose common mistakes and practise multi-step straight-line motion.”

Learning Objectives

• Choose the appropriate definitions or equations for a motion problem. • Check signs, units, intervals and assumptions. • Combine graph, algebraic and physical descriptions. • Explain the reasoning behind numerical and conceptual answers.

Build a description before calculating

Start by naming the object and reference frame. Choose an origin and positive axis direction, define the zero of time, and label the quantities you know. Decide whether the body can be treated as a point for this question. Then identify the motion phases: resting, constant velocity, constant acceleration, or a brief interaction such as a bounce. A correct model makes the calculation shorter and prevents inconsistent signs.

Quantity	Meaning or relation	SI unit and dimension
Position x	Coordinate relative to an origin	m; [L]
Displacement Δx	Final position minus initial position	m; [L]
Distance D	Total path length	m; [L]
Average velocity	Δx / Δt	m/s; [L T⁻¹]
Average speed	D / Δt	m/s; [L T⁻¹]
Instantaneous velocity	dx/dt	m/s; [L T⁻¹]
Instantaneous speed	\|v\|	m/s; [L T⁻¹]
Average acceleration	Δv / Δt	m/s²; [L T⁻²]
Instantaneous acceleration	dv/dt	m/s²; [L T⁻²]

A compact equation map

Constant accelerationLaTeX

v = u + a t, s = u t + \frac{1}{2} a t^{2}, v^{2} = u^{2} + 2 a s, s = \frac{u + v}{2} t

s = x − x₀ is displacement. These equations require constant acceleration over the selected interval.

General graph relationshipsLaTeX

v = \frac{d x}{d t}, a = \frac{d v}{d t}, Δ x = \int v d t, D = \int ∣ v ∣ d t

Integrate between the stated initial and final times. Signed velocity-time area gives displacement, not automatically distance.

Relative motionLaTeX

x_{B / A} = x_{B} - x_{A}, v_{B / A} = v_{B} - v_{A}

Use one common frame for the quantities being subtracted.

Physical situation	Useful result	Condition
Drop from rest	d = ½gt²; speed = gt	Negligible drag, constant g
Upward throw	t_up = u/g; H = u²/(2g)	u is positive upward launch speed
Return to launch height	T = 2u/g	Same release and landing heights
Uniform braking	d_b = u²/(2b); t_b = u/b	b is positive deceleration magnitude
Reaction travel	d_r = ut_r	Approximately constant speed during response
Ruler drop	t_r = √(2d/g)	Released from rest

Use the reasoning, not just a remembered formula

An equation is useful only if its assumptions match the interval. Check that all velocities refer to the same observer, that kilometres per hour have been converted if other quantities use metres and seconds, and that initial position is included. When a quadratic gives two times, inspect the physical interval rather than automatically accepting both. When taking a square root for velocity, restore direction from the trajectory.

Practice: a two-stage vehicle journey

Problem
A vehicle starts from rest, accelerates at 2 m/s² for 5 s, then continues at constant velocity for 10 s. Find total distance and average speed.

1.At the end of acceleration, v = 2 × 5 = 10 m/s. The first distance is ½ × 2 × 5² = 25 m.
2.The next phase covers 10 × 10 = 100 m. It has zero acceleration and must be handled separately.
3.Total distance = 125 m in 15 s, so average speed = 8.33 m/s. Motion never reverses, so with forward positive the average velocity is +8.33 m/s.

Practice: infer motion from a polynomial

Problem
Position is x = 4 + 12t − 3t² metres for 0 ≤ t ≤ 4 s. Find the turning time, displacement and distance.

1.Differentiate: v = 12 − 6t m/s and a = −6 m/s².
2.The turn occurs at t = 2 s. Positions are x(0) = 4 m, x(2) = 16 m and x(4) = 4 m.
3.Displacement is zero; distance is |16 − 4| + |4 − 16| = 24 m.
4.Average speed is 24/4 = 6 m/s and average velocity is zero. The nonzero initial position does not affect the displacement difference.

Practice: launch from above the ground

Problem
A ball is thrown downward at 5 m/s from a height of 30 m. Use g = 10 m/s². Find impact time and speed.

1.Choose downward positive and the launch point as origin. Then u = +5 m/s, a = +10 m/s² and s = +30 m.
2.Use 30 = 5t + 5t², giving (t + 3)(t − 2) = 0. The future impact occurs at t = 2 s.
3.Velocity is 5 + 10 × 2 = +25 m/s, so impact speed is 25 m/s. Check: 25² = 5² + 2 × 10 × 30.

Practice: check an average-speed shortcut

Problem
An object moves 100 m at 10 m/s and another 100 m at 20 m/s in the same direction. Is its average speed 15 m/s?

1.The times are 10 s and 5 s. Total time is 15 s.
2.Average speed is 200/15 = 13.33 m/s. The value 15 m/s would apply to equal durations, not these equal distances.
3.The smaller speed lasts longer, so it contributes more time weight. Because there is no reversal, the average velocity magnitude equals this average speed.

Practice: a delayed faster traveller

Problem
A walker starts at 1 m/s. A cyclist starts from the same point 10 s later at 6 m/s. When does the cyclist catch the walker?

1.Use t measured from the cyclist’s start. At this instant the walker has a 10 m lead.
2.The gap closes at 6 − 1 = 5 m/s, so catch time is 10/5 = 2 s after the cyclist starts.
3.This is 12 s after the walker starts, at a position 12 m from the common starting point. Both answers for time are correct only when their clock origins are stated.

Concept checks with explanations

Claim	Decision and reason
Negative acceleration always reduces speed	False: it increases speed when velocity is negative.
Zero speed can coexist with nonzero velocity	False: instantaneous speed equals \|v\|.
Zero velocity at an instant requires zero acceleration	False: the top of an upward throw is a counterexample.
Constant velocity means zero acceleration	True on a differentiable interval, because dv/dt = 0.
A curved x–t graph means a curved path	False: it can describe changing velocity along a straight line.
Distance equals \|displacement\| whenever endpoints differ	False: equality requires no reversal during the interval.
The standard three motion equations always hold	False: constant acceleration is required.
A speed graph has the same sign information as a velocity graph	False: speed omits direction.

Check your own solution

After calculating, ask whether the units match the requested quantity and whether the sign agrees with the stated direction. Compare distance with displacement magnitude: distance cannot be smaller. Check whether a claimed stopping time leaves velocity at zero, or whether a meeting time gives equal positions for both objects. For a vertical throw, verify that acceleration stays downward at the top. A rough graph often exposes a mistaken sign or an overlooked reversal before a long calculation does.

To practise independently, cover the worked solutions above, sketch each motion and solve it using a second representation. For a constant-acceleration calculation, check the result using the area of a velocity-time graph. For a relative-motion calculation, check the meeting with both separate position equations. Agreement between methods is more useful than memorising an answer.

Quick check

An object starts with v = +4 m/s and keeps a = −2 m/s² for 4 s. Which result is correct?

Key Takeaways

• Choose a frame, signs and interval before calculating. • Distinguish position, displacement and distance; distinguish average from instantaneous rates. • Read slopes and signed areas using the correct axes. • Use constant-acceleration formulas only where their assumptions hold. • Split at reversals for distance and at changes of acceleration for phase-by-phase calculations.

Previous · Lesson 9

Interpreting Complex Motion Graphs

End of chapter