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Motion and Graphs
Displacement, Velocity, and Acceleration
Displacement is a vector representing the straight-line change in position from initial to final point, while distance is the total path length traveled (a scalar).
$$\vec{d} = \vec{r_2} - \vec{r_1}$$
Displacement is the vector difference between final and initial position vectors.
$\vec{d}$=Displacement vector(m)
$\vec{r_1}$=Initial position vector(m)
$\vec{r_2}$=Final position vector(m)
Object returns to start
→Displacement = 0 even if distance traveled is large
Straight-line motion
→Displacement magnitude equals distance (path coincides with displacement vector)
Key Distinction: A car driving 5 km north then 5 km south has distance = 10 km but displacement = 0 km.
Scalar vs Vector: Distance is always positive; displacement can be negative (indicating direction).
Average velocity is total displacement divided by total time, while instantaneous velocity is the velocity at a specific instant — the limiting value of $\Delta d / \Delta t$ as $\Delta t \to 0$.
$$v_{avg} = \frac{\Delta d}{\Delta t}, \quad v_{inst} = \lim_{\Delta t \to 0} \frac{\Delta d}{\Delta t}$$
Average velocity gives an overall rate of position change; instantaneous velocity gives the rate at a single moment.
$v_{avg}$=Average velocity(m/s)
$v_{inst}$=Instantaneous velocity(m/s)
$\Delta d$=Change in displacement(m)
Uniform velocity
→$v_{avg} = v_{inst}$ at all times
Object returns to start
→$v_{avg} = 0$ regardless of how fast it moved
Graphical Meaning: On an s-t graph, average velocity = chord slope; instantaneous velocity = tangent slope.
Speed vs Velocity: Speed is the magnitude of velocity (always positive). Average speed = total distance / total time, which differs from average velocity when direction changes.
Limiting Case: As $\Delta t \to 0$, average velocity converges to instantaneous velocity. For uniform motion, they are always identical.
Acceleration is the time rate of change of velocity. It occurs whenever there is a change in speed, direction, or both.
$$a = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{\Delta t}$$
Acceleration is how quickly velocity changes — it is a vector in the direction of the velocity change.
$a$=Acceleration(m/s²)
$\Delta v$=Change in velocity(m/s)
$\Delta t$=Time interval(s)
$a > 0$ and $v > 0$
→Object speeds up in positive direction
$a < 0$ and $v > 0$
→Object slows down (deceleration)
$a$ and $v$ have same sign
→Speed increases; opposite signs means speed decreases
Uniform Acceleration: When velocity changes by equal amounts in equal time intervals — average acceleration equals instantaneous acceleration.
Dimensional Check: Acceleration has dimensions $[LT^{-2}]$. If your answer has units like m/s or m, you've made an error.
Interpreting Displacement-Time (s-t) Graphs
A displacement-time graph tracks an object's position relative to an origin; its gradient reveals the object's instantaneous velocity.
$$v = \frac{ds}{dt} \approx \frac{\Delta s}{\Delta t}$$
The slope of the tangent at any point on an s-t graph equals the velocity at that instant.
$v$=Velocity(m/s)
$\Delta s$=Change in displacement(m)
$\Delta t$=Time interval(s)
Graph is horizontal
→Object is stationary ($v = 0$)
Graph is a straight diagonal
→Object moves with constant velocity
Graph is curved
→Object is accelerating or decelerating
Gradient Significance: A steeper slope indicates higher speed; a negative slope indicates motion back toward the origin.
Instantaneous vs Average: The tangent slope gives instantaneous velocity, while the chord between two points gives average velocity.
Curvature: Upward concavity (curve getting steeper) indicates positive acceleration, while downward concavity (curve flattening) indicates deceleration.
Proportionality: For constant acceleration from rest, $s \propto t^2$ — doubling time quadruples displacement.
Velocity-Time (v-t) Graphs: Slope and Area
Velocity-time graphs describe how fast an object moves over time, where the slope represents acceleration and the area under curve represents displacement.
$$a = \frac{dv}{dt}, \quad \Delta s = \int v \, dt$$
The gradient determines the rate of velocity change, while the geometric area equates to the total displacement.
$a$=Acceleration(m/s²)
$\Delta v$=Change in velocity(m/s)
$\Delta s$=Total displacement(m)
Slope is zero (horizontal line)
→Uniform motion (constant velocity, zero acceleration)
Area is below time-axis
→Negative displacement (moving backwards)
Line crosses time-axis
→Object reverses direction at that instant
Acceleration vs Deceleration: A positive gradient implies speeding up in the positive direction; a negative gradient implies deceleration.
Total Distance vs Displacement: Total distance is the sum of absolute areas (all positive), while displacement is the algebraic sum of areas (positive minus negative).
Area Shapes: Triangle = $\frac{1}{2}bh$, Rectangle = $bh$, Trapezium = $\frac{1}{2}(a+b)h$ — these geometric formulas are essential for v-t graph problems.
Proportionality: $a \propto$ slope — doubling acceleration doubles the gradient of the v-t line.
Acceleration-Time (a-t) Graphs
Acceleration-time graphs show the rate of change of velocity; the area under this graph represents the change in velocity.
$$\Delta v = \text{Area under } a\text{-}t \text{ graph}$$
Integrating acceleration over a time period yields the total velocity gained or lost.
$\Delta v$=Change in velocity(m/s)
$a$=Acceleration(m/s²)
$a = 0$
→Velocity remains constant
Constant positive $a$
→v-t graph is a straight line with positive slope; s-t graph is a parabola
Constant Acceleration: Represented by a horizontal line on the a-t graph. This results in a linear v-t graph and a parabolic s-t graph.
Graph Chain: These three graph types are linked by differentiation (slope) and integration (area): s-t → (slope) → v-t → (slope) → a-t, and a-t → (area) → v-t → (area) → s-t.
Limiting Case: For free fall, the a-t graph is a horizontal line at $a = g = 9.8$ m/s² (or $-9.8$ m/s² if upward is positive).
Equations of Uniformly Accelerated Motion
For motion with constant acceleration along a straight line, four kinematic equations relate displacement, velocity, acceleration, and time.
First Equation: $v_f = v_i + at$ — gives final velocity from initial velocity and acceleration.
Second Equation: $s = \frac{v_i + v_f}{2} \times t$ — displacement from average of initial and final velocities.
Third Equation: $s = v_i t + \frac{1}{2}at^2$ — displacement without knowing final velocity.
Fourth Equation: $v_f^2 = v_i^2 + 2as$ — final velocity without knowing time.
Choosing the Right Equation: Identify which variable is missing from the problem, then use the equation that does not contain it.
Equation Selection Guide
Pick the equation based on the unknown you do NOT need
•
Missing $s$: use $v_f = v_i + at$
•
Missing $a$: use $s = \frac{v_i + v_f}{2} \times t$
•
Missing $v_f$: use $s = v_i t + \frac{1}{2}at^2$
•
Missing $t$: use $v_f^2 = v_i^2 + 2as$
The distance covered in the $n^{th}$ second of motion (not the first $n$ seconds) is given by a special formula derived from the third equation of motion.
$$s_n = v_i + \frac{a}{2}(2n - 1)$$
This gives the displacement in a single specific second, not the cumulative displacement.
$s_n$=Distance in the nth second(m)
$v_i$=Initial velocity(m/s)
$a$=Constant acceleration(m/s²)
$n$=The specific second (1st, 2nd, 3rd...)(dimensionless)
$v_i = 0$ (starting from rest)
→Distances in successive seconds follow the odd number ratio 1 : 3 : 5 : 7...
$a = 0$ (uniform velocity)
→Distance in every second is the same: $s_n = v_i$
Derivation Sketch: $s_n = s(n) - s(n-1) = [v_i n + \frac{1}{2}an^2] - [v_i(n-1) + \frac{1}{2}a(n-1)^2]$, which simplifies to the formula above.
Proportionality: For $v_i = 0$, the distance in the $n^{th}$ second $\propto (2n - 1)$. This is linear in $n$, not quadratic.
In the absence of air resistance, all objects near the Earth's surface fall with a uniform acceleration $g \approx 9.8$ m/s², regardless of their mass.
Sign Convention: Choose upward or downward as positive and stay consistent. If upward is positive, then $a = -g = -9.8$ m/s².
Symmetry: Time going up = time coming down (same height). Speed at launch = speed on return (same height).
Proportionality: For an object dropped from rest, $s \propto t^2$ — falling 4× the distance in 2× the time.
Application Strategy
Worked Examples
Problem Statement
An athlete runs 100 m. The s-t graph is a parabola for the first 4 seconds (accelerating from rest) and then a straight line for the remaining 6 seconds. If the athlete covers 100 m total in 10 s, find the velocity during the final 6 seconds.
Slope of s-t graphUniform velocityArea under parabola
Strategic Toolkit
$$v = \frac{\Delta s}{\Delta t}$$
Step 1
Analyze Motion Phases
The first 4 s is acceleration from rest ($s \propto t^2$). The final 6 s is constant velocity ($s \propto t$). At the transition point ($t = 4$ s), the velocity is $v$.
Step 2
Establish Relationship
For constant acceleration from rest, the s-t graph is a parabola. The distance in the first phase: $s_1 = \frac{1}{2}v \times 4 = 2v$ (area of triangle on the v-t graph). The remaining distance in the constant velocity phase: $s_2 = v \times 6 = 6v$.
$s_1 + s_2 = 100 \implies 2v + 6v = 100$
Step 3
Solve for Velocity
Combining the two phases:
$8v = 100 \implies v = 12.5$ m/s
Problem Statement
A car's velocity changes from 0 to 20 m/s in 5 seconds, stays constant for 10 seconds, and then brakes to a stop in 2 seconds. Find the total distance traveled.
Area under v-t graphTrapezium areaGeometric decomposition
Strategic Toolkit
$$\text{Distance} = \text{Area under } v\text{-}t$$
Step 1
Divide into Geometries
The graph consists of a triangle (0–5 s), a rectangle (5–15 s), and another triangle (15–17 s).
Step 2
Calculate Individual Areas
Use $\frac{1}{2}bh$ for triangles and $lw$ for the rectangle.
$A_1 = \frac{1}{2} \times 5 \times 20 = 50$ m
$A_2 = 10 \times 20 = 200$ m
$A_3 = \frac{1}{2} \times 2 \times 20 = 20$ m
Step 3
Sum Areas
The total distance is the sum of the areas of all three phases.
$S_{total} = 50 + 200 + 20 = 270$ m
Problem Statement
A stone is dropped from rest from the top of a 45 m building. Find (a) the time to reach the ground, and (b) the velocity just before impact. Take $g = 10$ m/s².
Equations of motionFree fallChoosing the right equation
Strategic Toolkit
$$s = v_i t + \frac{1}{2}at^2, \quad v_f^2 = v_i^2 + 2as$$
Step 1
Identify Known Quantities
Taking downward as positive: $v_i = 0$ (dropped from rest), $a = g = 10$ m/s², $s = 45$ m.
Step 2
Find Time (Part a)
Use $s = v_i t + \frac{1}{2}at^2$. Since $v_i = 0$:
$45 = 0 + \frac{1}{2}(10)t^2 \implies t^2 = 9 \implies t = 3$ s
Step 3
Find Final Velocity (Part b)
Use $v_f^2 = v_i^2 + 2as$:
$v_f^2 = 0 + 2(10)(45) = 900 \implies v_f = 30$ m/s
Problem Statement
An object starts from the origin. Its v-t graph shows 4 m/s for 3 s, then 0 m/s for 2 s, then −2 m/s for 3 s. Sketch the corresponding s-t graph and find the final displacement.
Graph conversionArea under v-t graphDisplacement vs distance
Strategic Toolkit
$$\Delta s = v \times \Delta t$$
Step 1
Phase 1 (0–3 s): Constant positive velocity
Displacement = $4 \times 3 = 12$ m. The s-t graph rises linearly from 0 to 12 m.
Step 2
Phase 2 (3–5 s): Stationary
Displacement = $0 \times 2 = 0$ m. The s-t graph is horizontal at 12 m.
Step 3
Phase 3 (5–8 s): Constant negative velocity
Displacement = $(-2) \times 3 = -6$ m. The s-t graph decreases linearly from 12 m to 6 m.
Final displacement = $12 + 0 + (-6) = 6$ m
Total distance = $12 + 0 + 6 = 18$ m
Problem Statement
An object starts from rest with uniform acceleration of 4 m/s². Find the distance traveled in the 5th second of motion.
nth second formulaOdd number ratio
Strategic Toolkit
$$s_n = v_i + \frac{a}{2}(2n - 1)$$
Step 1
Identify Known Quantities
$v_i = 0$ (starts from rest), $a = 4$ m/s², $n = 5$.
Step 2
Apply the Formula
Substitute into $s_n = v_i + \frac{a}{2}(2n - 1)$:
$s_5 = 0 + \frac{4}{2}(2 \times 5 - 1) = 2 \times 9 = 18$ m
Step 3
Verify with Subtraction
Cross-check: total distance in 5 s minus total in 4 s.
$s(5) = \frac{1}{2}(4)(25) = 50$ m, $s(4) = \frac{1}{2}(4)(16) = 32$ m. Difference = $50 - 32 = 18$ m ✓
Problem Statement
A car traveling at 72 km/h brakes with a uniform deceleration of 5 m/s². How far does it travel before stopping?
Unit conversionFourth equationDeceleration
Strategic Toolkit
$$v_f^2 = v_i^2 + 2as$$
Step 1
Convert Units
Convert 72 km/h to m/s:
$v_i = \frac{72}{3.6} = 20$ m/s
Step 2
Set Up the Equation
Since the car stops, $v_f = 0$. Deceleration means $a = -5$ m/s².
$0 = 20^2 + 2(-5)s \implies 10s = 400$
Step 3
Solve
Solving for $s$:
$s = \frac{400}{10} = 40$ m
Memorize formulas
Gradient of s-t
The slope of a displacement-time graph gives the velocity.
Formula
$v = \frac{ds}{dt}$
Gradient of v-t
The slope of a velocity-time graph gives the acceleration.
Formula
$a = \frac{dv}{dt}$
Area under v-t
The area between the velocity curve and the time axis gives the displacement.
Formula
$\Delta s = \int_{t_1}^{t_2} v \, dt$
Area under a-t
The area under an acceleration-time graph gives the change in velocity.
Formula
$\Delta v = \int_{t_1}^{t_2} a \, dt$
First Equation of Motion
Final velocity from initial velocity, acceleration, and time.
Formula
$v_f = v_i + at$
Second Equation of Motion
Displacement from average velocity and time.
Formula
$s = \frac{v_i + v_f}{2} \times t$
Third Equation of Motion
Displacement without knowing final velocity.
Formula
$s = v_i t + \frac{1}{2}at^2$
Fourth Equation of Motion
Final velocity without knowing time.
Formula
$v_f^2 = v_i^2 + 2as$
Distance in nth Second
Displacement covered in a specific second of uniformly accelerated motion.
Formula
$s_n = v_i + \frac{a}{2}(2n - 1)$
Review
1
A displacement-time graph shows a curve that is getting flatter (gradient decreasing) over time in the positive direction. What does this describe?
Ans
The object is decelerating while moving in the positive direction.
Review
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Read the slope as velocity
For a displacement-time graph, the gradient gives the velocity. A positive slope means forward motion, and a smaller slope means the velocity is getting smaller.
Positive direction with decreasing gradient = slowing down while still moving forward.
2
A v-t graph shows an object starting at 8 m/s and uniformly increasing to 20 m/s over 6 seconds. What is the displacement during this interval?
Ans
84 m
Review
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Use area under the velocity-time graph
Displacement equals the area under a v-t graph. Here the shape is a trapezium with parallel sides 8 and 20, and width 6 s.
$$s = \frac{1}{2}(8 + 20)\times 6 = 84\ \text{m}$$
3
An object starts from rest and accelerates uniformly. The distances covered in the 1st, 2nd, and 3rd seconds are in the ratio:
Ans
1 : 3 : 5
Review
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Apply the nth-second pattern
For motion from rest with uniform acceleration, the distance covered in each successive second goes as odd numbers.
$$s_1:s_2:s_3 = 1:3:5$$
4
A ball is thrown vertically upward with initial velocity 20 m/s. Taking $g = 10$ m/s² and upward as positive, what is its velocity after 3 seconds?
Ans
−10 m/s
Review
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Use the first equation of motion
Take upward as positive, so the acceleration due to gravity is $-10$ m/s². Substitute the given values into $v = u + at$.
$$v = 20 + (-10)(3) = -10\ \text{m/s}$$
5
Which feature of a velocity-time graph represents the instantaneous acceleration at a specific point in time?
Ans
The gradient of the tangent at that point.
Review
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Use the slope-derivative link
Instantaneous acceleration is the rate of change of velocity, so on a v-t graph it is given by the slope at that point, not the area or intercept.
Acceleration = gradient of the tangent.
6
An object moving at 10 m/s has its a-t graph showing a constant acceleration of 2 m/s² for 5 seconds, then a deceleration of −4 m/s² for 5 seconds. What is the velocity at $t = 10$ s?
Ans
0 m/s
Review
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Find the velocity change in each phase
Acceleration changes velocity by the area under the a-t graph. Over the first 5 s the velocity increases by $2\times 5$, then over the next 5 s it decreases by $4\times 5$.
$$v = 10 + (2\times 5) + (-4\times 5) = 0\ \text{m/s}$$
Check the result against both intervals
After the first interval the velocity is 20 m/s, and the second interval removes 20 m/s of velocity. That confirms the object comes to rest at 10 s.
Final velocity at $t=10$ s = $0$ m/s
7
A car traveling at 72 km/h applies brakes and comes to rest in 50 m. What is the magnitude of its deceleration?
Ans
4 m/s²
Review
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Convert the speed to SI units
Use the standard kinematics equations in m/s and m. Convert 72 km/h to 20 m/s before substituting.
$$72\ \text{km/h} = 20\ \text{m/s}$$
Apply the velocity-distance relation
With final speed zero and stopping distance 50 m, use $v^2 = u^2 + 2as$ and solve for $a$.
$$0 = 20^2 + 2a(50) \Rightarrow a = -4\ \text{m/s}^2$$
8
Two objects are dropped from heights $h$ and $4h$ respectively. The ratio of their times to reach the ground is:
Ans
1 : 2
Review
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Use free-fall time dependence
For an object dropped from rest, $h = \frac{1}{2}gt^2$, so time is proportional to the square root of height.
$$t \propto \sqrt{h} \Rightarrow \frac{t_1}{t_2} = \sqrt{\frac{h}{4h}} = \frac{1}{2}$$
9
On a v-t graph, a straight line starts at $v = 15$ m/s and reaches $v = 0$ in 5 seconds. What is the displacement?
Ans
37.5 m
Review
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Treat the graph area as a triangle
Displacement is the area under the v-t graph. Since velocity falls linearly from 15 to 0 in 5 s, the area is triangular.
$$s = \frac{1}{2}\times 5 \times 15 = 37.5\ \text{m}$$
10
A body is projected vertically upward. At the highest point, which of the following is true?
Ans
Velocity is zero but acceleration is $g$ downward.
Review
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Separate velocity from acceleration
At the peak, the body momentarily stops, so the velocity is zero. Gravity still acts, so the acceleration remains downward with magnitude $g$.
Velocity = 0, acceleration = $g$ downward.
11
A runner jogs at 4 m/s for 20 seconds, then sprints at 8 m/s for 10 seconds. What is the average speed for the entire run?
Ans
5.33 m/s
Review
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Compute the total distance first
Average speed must use total distance divided by total time, so find the distance covered in each interval and add them.
$$d = 4(20) + 8(10) = 160\ \text{m}$$
Divide by the total time
The full time is 30 s, so the average speed is the total distance divided by 30.
$$v_{avg} = \frac{160}{30} = 5.33\ \text{m/s}$$
12
The s-t graph of an object is a parabola opening downward. At the vertex (peak) of the parabola, the object's velocity is:
Ans
Zero
Review
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Read the tangent slope at the peak
Velocity on an s-t graph is the slope. At the vertex of a downward parabola the tangent is horizontal, so the slope is zero.
Velocity at the vertex = 0
13
An object starts from rest with uniform acceleration $a$. The ratio of the distance covered in the 3rd second to the distance covered in the 5th second is:
Ans
5 : 9
Review
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Use the odd-number pattern for nth-second distance
For motion from rest with constant acceleration, the distance in the nth second is proportional to $2n-1$.
$$s_3:s_5 = (2\cdot3-1):(2\cdot5-1) = 5:9$$
14
A car covers one-third of a distance at 20 km/h, the next third at 30 km/h, and the last third at 60 km/h. The average speed for the whole journey is:
Ans
30 km/h
Review
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Convert equal-distance segments into times
Because each third of the journey has the same distance, add the times for each segment instead of averaging the speeds directly.
$$t = \frac{d}{20} + \frac{d}{30} + \frac{d}{60} = \frac{d}{10}$$
Compute the average speed
Average speed is total distance divided by total time. The total distance is $3d$, so divide by the total time found above.
$$v_{avg} = \frac{3d}{d/10} = 30\ \text{km/h}$$
15
A ball is thrown upward with velocity $u$. Taking upward as positive, which pair of graphs correctly represents the motion? (s-t and v-t)
Ans
s-t: downward parabola; v-t: straight line with negative slope
Review
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Match each graph to the motion law
With constant downward acceleration, velocity changes linearly with time, so the v-t graph is a straight line with negative slope. Displacement therefore follows a parabola opening downward.
s-t: downward parabola; v-t: straight line with negative slope.
Question 1 of 15
0:00
1
A displacement-time graph shows a curve that is getting flatter (gradient decreasing) over time in the positive direction. What does this describe?