The Space Under a Line Tells You How Far the Car Went
Learn to read speed from the slope of a distance-time graph, find acceleration and distance from a velocity-time graph, sketch both graphs for four kinds of motion, and derive the equations of motion from a graph.
Why does the area under a velocity-time graph give a distance?
Draw a velocity-time graph for a car moving at a steady for s. It is a horizontal line at height , and the region under it is a rectangle.
That rectangle's area is . And the distance the car covered is m.
The same number, and not by accident. Look at the units of the two sides you multiplied:
The seconds cancel and a length comes out. So the "area" of that region is not an area at all — it is a distance, because the two axes carry velocity and time rather than two lengths.
That single observation makes a velocity-time graph enormously useful. Its slope gives the acceleration and the region under it gives the distance, so one picture answers both questions no matter how complicated the motion. And the last section of this page shows that the equations of motion themselves drop out of the same graph.
This page covers the second part of the ICSE Class 9 Physics chapter on motion in one dimension — reading distance-time and velocity-time graphs, sketching them, and deriving the equations graphically.
That rectangle's area is . And the distance the car covered is m.
The same number, and not by accident. Look at the units of the two sides you multiplied:
The seconds cancel and a length comes out. So the "area" of that region is not an area at all — it is a distance, because the two axes carry velocity and time rather than two lengths.
That single observation makes a velocity-time graph enormously useful. Its slope gives the acceleration and the region under it gives the distance, so one picture answers both questions no matter how complicated the motion. And the last section of this page shows that the equations of motion themselves drop out of the same graph.
This page covers the second part of the ICSE Class 9 Physics chapter on motion in one dimension — reading distance-time and velocity-time graphs, sketching them, and deriving the equations graphically.
How do you read speed from a distance-time graph?
The slope of a distance-time graph is the speed — how much distance is gained per second.
What the shapes mean.
- A horizontal line — the distance is not changing, so the body is at rest
- A straight sloping line — equal distances in equal times, so uniform motion at constant speed
- A curve — the distance gained per second is changing, so non-uniform motion
- A steeper line means a greater speed
Worked example 1. A graph rises in a straight line from the origin to the point :
Worked example 2 — not starting at the origin. A straight line runs from to :
The body began m from the reference point, and that starting offset does not affect the speed. **Dividing by would give , which is the commonest error here — the slope needs the change in distance, not the final reading.
Worked example 3 — a journey in three stages.** A graph rises from to , stays flat until s, then rises to .
- Stage 1:
- Stage 2: slope zero, so at rest for s
- Stage 3:
Average speed for the whole journey:
which is less than either moving speed, because of the stop in the middle.
Worked example 4 — a curve. A graph bending upward means the slope is increasing, so the body is speeding up. Bending towards the horizontal means the slope is decreasing, so it is slowing down.
A distance-time graph can never slope downwards. Distance is a total path length and cannot decrease — the body may stop, giving a horizontal line, but it can never un-travel. A graph that falls must be a displacement-time graph, where a fall means moving back towards the starting point. That is a genuine distinction between the two graph types, and reading the axis label is what settles which one you have.
And a distance-time graph can never be vertical. A vertical segment would mean covering distance in no time at all, which no body does.
What the shapes mean.
- A horizontal line — the distance is not changing, so the body is at rest
- A straight sloping line — equal distances in equal times, so uniform motion at constant speed
- A curve — the distance gained per second is changing, so non-uniform motion
- A steeper line means a greater speed
Worked example 1. A graph rises in a straight line from the origin to the point :
Worked example 2 — not starting at the origin. A straight line runs from to :
The body began m from the reference point, and that starting offset does not affect the speed. **Dividing by would give , which is the commonest error here — the slope needs the change in distance, not the final reading.
Worked example 3 — a journey in three stages.** A graph rises from to , stays flat until s, then rises to .
- Stage 1:
- Stage 2: slope zero, so at rest for s
- Stage 3:
Average speed for the whole journey:
which is less than either moving speed, because of the stop in the middle.
Worked example 4 — a curve. A graph bending upward means the slope is increasing, so the body is speeding up. Bending towards the horizontal means the slope is decreasing, so it is slowing down.
A distance-time graph can never slope downwards. Distance is a total path length and cannot decrease — the body may stop, giving a horizontal line, but it can never un-travel. A graph that falls must be a displacement-time graph, where a fall means moving back towards the starting point. That is a genuine distinction between the two graph types, and reading the axis label is what settles which one you have.
And a distance-time graph can never be vertical. A vertical segment would mean covering distance in no time at all, which no body does.
Formula
How do you get acceleration and distance from a velocity-time graph?
The slope gives the acceleration and the region under the line gives the distance:
Worked example 1 — uniform acceleration. A straight line runs from to .
The slope gives the acceleration:
The region under it is a trapezium with parallel sides and and width :
Check with the equation of motion: m. The same answer, and the agreement is not a coincidence — the last section shows the equation comes from this area.
Worked example 2 — uniform velocity. A horizontal line at for s. Slope zero, so . Area is a rectangle:
Worked example 3 — retardation to rest. A line falls from to .
a retardation of . The region is a triangle:
Check: m. Correct.
Worked example 4 — accelerating from rest. A line from the origin to :
Worked example 5 — a three-phase journey. A bus accelerates from rest to in s, holds that speed for s, then brakes to rest in s. Split the region into a triangle, a rectangle and a triangle:
- accelerating: m
- steady: m
- braking: m
The total time is s, so
Splitting the region into standard shapes is the whole method, and it works however many phases a journey has.
A velocity-time graph CAN slope downwards, and can even go below the axis. Falling towards the axis means slowing down; crossing below it means the body has reversed direction. Area below the axis counts as negative displacement, so for a body that goes out and comes back the areas partly cancel — which is why the region gives the displacement in general and the distance only when the motion never reverses.
Worked example 1 — uniform acceleration. A straight line runs from to .
The slope gives the acceleration:
The region under it is a trapezium with parallel sides and and width :
Check with the equation of motion: m. The same answer, and the agreement is not a coincidence — the last section shows the equation comes from this area.
Worked example 2 — uniform velocity. A horizontal line at for s. Slope zero, so . Area is a rectangle:
Worked example 3 — retardation to rest. A line falls from to .
a retardation of . The region is a triangle:
Check: m. Correct.
Worked example 4 — accelerating from rest. A line from the origin to :
Worked example 5 — a three-phase journey. A bus accelerates from rest to in s, holds that speed for s, then brakes to rest in s. Split the region into a triangle, a rectangle and a triangle:
- accelerating: m
- steady: m
- braking: m
The total time is s, so
Splitting the region into standard shapes is the whole method, and it works however many phases a journey has.
A velocity-time graph CAN slope downwards, and can even go below the axis. Falling towards the axis means slowing down; crossing below it means the body has reversed direction. Area below the axis counts as negative displacement, so for a body that goes out and comes back the areas partly cancel — which is why the region gives the displacement in general and the distance only when the motion never reverses.
How do you sketch the graphs for four kinds of motion?
Decide what the slope must do in each case, then draw the shape that has that slope. The distance-time and velocity-time pictures always come in pairs.
A body at rest.
- Distance-time: a horizontal line — the distance never changes
- Velocity-time: a horizontal line along the time axis, at
Uniform velocity.
- Distance-time: a straight sloping line. For it passes through , , — equal steps for equal times
- Velocity-time: a horizontal line above the axis, at the constant value
Uniform acceleration.
- Distance-time: a curve bending upward, because the distance gained per second keeps growing. Starting from rest with , the readings are , giving m, m, m, m at , , and s — differences of , , , m, which grow
- Velocity-time: a straight line rising from , with slope
Uniform retardation.
- Distance-time: a curve flattening out, because each second adds less than the one before. It becomes horizontal at the instant the body stops
- Velocity-time: a straight line falling towards the axis
Worked example — reading a value off a sketch. For a body starting from rest at , the velocity-time graph is a straight line of slope through the origin. At s it reads , and the triangle under it has area
Check against the distance-time table above: at gives m. The two graphs agree, as they must, since they describe one motion.
A straight line on one graph does not mean a straight line on the other. Uniform acceleration gives a straight velocity-time graph and a curved distance-time graph. Uniform velocity gives a straight distance-time graph and a flat velocity-time graph. So a question asking which graph is a straight line is asking which quantity is changing at a steady rate, and confusing the two pictures is the standard error in sketch questions.
Always label both axes with the quantity and its unit, and mark the scale. An unlabelled sketch cannot be read, and the same shape means quite different things on the two graphs.
A body at rest.
- Distance-time: a horizontal line — the distance never changes
- Velocity-time: a horizontal line along the time axis, at
Uniform velocity.
- Distance-time: a straight sloping line. For it passes through , , — equal steps for equal times
- Velocity-time: a horizontal line above the axis, at the constant value
Uniform acceleration.
- Distance-time: a curve bending upward, because the distance gained per second keeps growing. Starting from rest with , the readings are , giving m, m, m, m at , , and s — differences of , , , m, which grow
- Velocity-time: a straight line rising from , with slope
Uniform retardation.
- Distance-time: a curve flattening out, because each second adds less than the one before. It becomes horizontal at the instant the body stops
- Velocity-time: a straight line falling towards the axis
Worked example — reading a value off a sketch. For a body starting from rest at , the velocity-time graph is a straight line of slope through the origin. At s it reads , and the triangle under it has area
Check against the distance-time table above: at gives m. The two graphs agree, as they must, since they describe one motion.
A straight line on one graph does not mean a straight line on the other. Uniform acceleration gives a straight velocity-time graph and a curved distance-time graph. Uniform velocity gives a straight distance-time graph and a flat velocity-time graph. So a question asking which graph is a straight line is asking which quantity is changing at a steady rate, and confusing the two pictures is the standard error in sketch questions.
Always label both axes with the quantity and its unit, and mark the scale. An unlabelled sketch cannot be read, and the same shape means quite different things on the two graphs.
How do you derive v = u + at and S = ut + half a t squared from a graph?
Read the first equation off the slope and the second off the area. Both come from one straight-line velocity-time graph.
The setup. A body has initial velocity and uniform acceleration . On a velocity-time graph this is a straight line from the point to the point , where is the velocity after time .
First equation, from the slope. The slope of a velocity-time graph is the acceleration:
Multiplying through by and rearranging:
Second equation, from the area. The region under is a trapezium with parallel sides and and width :
That is already a usable result. Now substitute from the first equation:
Where the two terms come from on the picture. Split the trapezium into a rectangle and a triangle:
- the rectangle of height and width has area — the distance the body would have covered at its starting speed alone
- the triangle on top has base and height , so its area is — the extra distance the acceleration contributed
**So is not one formula but two pieces of the same region, and knowing which piece is which makes the equation impossible to misremember.
The third equation follows too.** Take and substitute from the first equation:
using the difference-of-squares identity, and rearranging:
Worked check on all three. A body starts at with for s.
-
- m
- , so
All three agree, and they match the trapezium area of m from the second section.
These equations hold only for UNIFORM acceleration. The derivation needed the velocity-time graph to be a straight line — that is what made the region a trapezium and the slope a single number. For a curved velocity-time graph the slope changes from point to point, the region is not a trapezium, and none of the three equations applies. So every problem using them must first be checked for constant acceleration, and the third part of this chapter applies them to exactly that case.
The setup. A body has initial velocity and uniform acceleration . On a velocity-time graph this is a straight line from the point to the point , where is the velocity after time .
First equation, from the slope. The slope of a velocity-time graph is the acceleration:
Multiplying through by and rearranging:
Second equation, from the area. The region under is a trapezium with parallel sides and and width :
That is already a usable result. Now substitute from the first equation:
Where the two terms come from on the picture. Split the trapezium into a rectangle and a triangle:
- the rectangle of height and width has area — the distance the body would have covered at its starting speed alone
- the triangle on top has base and height , so its area is — the extra distance the acceleration contributed
**So is not one formula but two pieces of the same region, and knowing which piece is which makes the equation impossible to misremember.
The third equation follows too.** Take and substitute from the first equation:
using the difference-of-squares identity, and rearranging:
Worked check on all three. A body starts at with for s.
-
- m
- , so
All three agree, and they match the trapezium area of m from the second section.
These equations hold only for UNIFORM acceleration. The derivation needed the velocity-time graph to be a straight line — that is what made the region a trapezium and the slope a single number. For a curved velocity-time graph the slope changes from point to point, the region is not a trapezium, and none of the three equations applies. So every problem using them must first be checked for constant acceleration, and the third part of this chapter applies them to exactly that case.
Exam tip
Exam tip: take the change, not the final reading
Slope means the CHANGE divided by the change. A line from to has slope , never .
Distance-time slope is speed; velocity-time slope is acceleration; velocity-time area is distance. Write which one you are using.
Check the area's units: . That confirms the region is a distance and not an area.
Split the region into triangles and rectangles and list each piece: m.
A distance-time graph can never fall or be vertical. A falling graph must be displacement-time.
Area below the velocity axis is negative displacement, so a there-and-back journey partly cancels.
Uniform acceleration gives a STRAIGHT velocity-time graph and a CURVED distance-time graph — do not draw the same shape twice.
Label both axes with quantity and unit, mark the scale, and use a ruler for straight portions.
For the derivations, quote the reason at each step: slope of a velocity-time graph is acceleration, area under it is distance. Marks are for the reasons.
Show the trapezium split as a rectangle plus a triangle — that is the part examiners look for.
And state that the equations need uniform acceleration; the derivation used a straight line and nothing else works.
Distance-time slope is speed; velocity-time slope is acceleration; velocity-time area is distance. Write which one you are using.
Check the area's units: . That confirms the region is a distance and not an area.
Split the region into triangles and rectangles and list each piece: m.
A distance-time graph can never fall or be vertical. A falling graph must be displacement-time.
Area below the velocity axis is negative displacement, so a there-and-back journey partly cancels.
Uniform acceleration gives a STRAIGHT velocity-time graph and a CURVED distance-time graph — do not draw the same shape twice.
Label both axes with quantity and unit, mark the scale, and use a ruler for straight portions.
For the derivations, quote the reason at each step: slope of a velocity-time graph is acceleration, area under it is distance. Marks are for the reasons.
Show the trapezium split as a rectangle plus a triangle — that is the part examiners look for.
And state that the equations need uniform acceleration; the derivation used a straight line and nothing else works.
Did you know
Why the odd numbers hide inside a falling stone's graph
Take a body starting from rest with a uniform acceleration of , and work out how far it has gone after each second using :
- after s: m
- after s: m
- after s: m
- after s: m
- after s: m
Now ask a different question: how far did it travel during each separate second? Subtract consecutive readings:
Divide through by the first one and the pattern is unmistakable: . The distances covered in successive seconds are in the ratio of the odd numbers, for any body starting from rest with any uniform acceleration.
The velocity-time graph shows why. It is a straight line through the origin, and the distance in each second is the area of the strip above that second. The first strip is a triangle; every later strip is a trapezium one step taller than the last. Each new strip gains the same fixed amount of height, and its area grows by two units of the first triangle's area every time — which produces exactly the odd numbers.
The sequences chapter met the same pattern from the other side: the sum of the first odd numbers is . Here that identity is doing physical work. Adding the strips m must equal the total at , and it does — because .
So a fact about odd numbers and a fact about falling stones are the same fact. The **distance grows as **, and squares grow by odd numbers, and that is all the ratio is.
- after s: m
- after s: m
- after s: m
- after s: m
- after s: m
Now ask a different question: how far did it travel during each separate second? Subtract consecutive readings:
Divide through by the first one and the pattern is unmistakable: . The distances covered in successive seconds are in the ratio of the odd numbers, for any body starting from rest with any uniform acceleration.
The velocity-time graph shows why. It is a straight line through the origin, and the distance in each second is the area of the strip above that second. The first strip is a triangle; every later strip is a trapezium one step taller than the last. Each new strip gains the same fixed amount of height, and its area grows by two units of the first triangle's area every time — which produces exactly the odd numbers.
The sequences chapter met the same pattern from the other side: the sum of the first odd numbers is . Here that identity is doing physical work. Adding the strips m must equal the total at , and it does — because .
So a fact about odd numbers and a fact about falling stones are the same fact. The **distance grows as **, and squares grow by odd numbers, and that is all the ratio is.
Exam relevance
How are motion graphs tested in JEE Main and NEET?
Because graph reading is faster than algebra for many problems, and the slope-and-area pair becomes the derivative-and-integral pair in Class 11.
This is the foundation for Class 11 Physics Motion in a Straight Line, examined in JEE Main and NEET. The two operations used on this page are the two operations of calculus in disguise:
So a student who reads a slope as a velocity and a region as a distance already has the physical meaning of the derivative and the integral, and Class 11 supplies only the notation and the ability to handle curves. The equations derived here graphically are re-derived there by integration, and the restriction to uniform acceleration becomes visible as the condition that lets the integral be done without calculus.
Graph-based questions are a recurring type in both exams. Typical forms are: given a velocity-time graph with several phases, find the total distance, the displacement, the average velocity, or the acceleration in a stated interval. The three-phase worked example on this page is exactly that question, and the split-into-shapes method solves it without any formula.
The distance-against-displacement distinction becomes the sign of the area. Class 11 asks for both from a graph that dips below the axis: the distance adds the magnitudes of all the regions, while the displacement subtracts the ones below. That single difference is a standard trap, and it is the point flagged on this page.
Acceleration-time graphs are added in Class 11, where the area under an acceleration-time graph gives the change in velocity — the same slope-and-area logic moved one step along. Questions chaining all three graphs together appear in JEE Main.
The odd-number ratio from the previous section reappears as a shortcut: for a body starting from rest, the distances in successive equal intervals go as , and recognising it turns some numericals into mental arithmetic. The third part of this chapter uses the same result as the th-second formula.
What the questions look like. For board work, expect find the speed or acceleration from a given graph, find the distance from the area, sketch the pair of graphs for a described motion, and **derive and graphically — the derivation is asked in words and needs the labelled graph plus a reason at each step. For JEE Main and NEET, expect multi-phase graph reading, sign-sensitive displacement questions, and matching a graph to a description.
How board and competitive emphasis differ. A board paper rewards the labelled graph and the stated reason for each step of the derivation. A competitive paper gives the graph and tests only whether you can extract the right number quickly — often by recognising a triangle's area rather than substituting into a formula.
The single trap that costs the most marks.** Using the final reading instead of the change when finding a slope. A distance-time line from to has a speed of , and dividing by gives a plausible . The defence is to write the subtraction explicitly — — so the starting offset cannot be quietly dropped.
This is the foundation for Class 11 Physics Motion in a Straight Line, examined in JEE Main and NEET. The two operations used on this page are the two operations of calculus in disguise:
So a student who reads a slope as a velocity and a region as a distance already has the physical meaning of the derivative and the integral, and Class 11 supplies only the notation and the ability to handle curves. The equations derived here graphically are re-derived there by integration, and the restriction to uniform acceleration becomes visible as the condition that lets the integral be done without calculus.
Graph-based questions are a recurring type in both exams. Typical forms are: given a velocity-time graph with several phases, find the total distance, the displacement, the average velocity, or the acceleration in a stated interval. The three-phase worked example on this page is exactly that question, and the split-into-shapes method solves it without any formula.
The distance-against-displacement distinction becomes the sign of the area. Class 11 asks for both from a graph that dips below the axis: the distance adds the magnitudes of all the regions, while the displacement subtracts the ones below. That single difference is a standard trap, and it is the point flagged on this page.
Acceleration-time graphs are added in Class 11, where the area under an acceleration-time graph gives the change in velocity — the same slope-and-area logic moved one step along. Questions chaining all three graphs together appear in JEE Main.
The odd-number ratio from the previous section reappears as a shortcut: for a body starting from rest, the distances in successive equal intervals go as , and recognising it turns some numericals into mental arithmetic. The third part of this chapter uses the same result as the th-second formula.
What the questions look like. For board work, expect find the speed or acceleration from a given graph, find the distance from the area, sketch the pair of graphs for a described motion, and **derive and graphically — the derivation is asked in words and needs the labelled graph plus a reason at each step. For JEE Main and NEET, expect multi-phase graph reading, sign-sensitive displacement questions, and matching a graph to a description.
How board and competitive emphasis differ. A board paper rewards the labelled graph and the stated reason for each step of the derivation. A competitive paper gives the graph and tests only whether you can extract the right number quickly — often by recognising a triangle's area rather than substituting into a formula.
The single trap that costs the most marks.** Using the final reading instead of the change when finding a slope. A distance-time line from to has a speed of , and dividing by gives a plausible . The defence is to write the subtraction explicitly — — so the starting offset cannot be quietly dropped.
Key takeaways
Motion graphs and the graphical derivations: quick revision
- Distance-time graph: the slope is the speed. Horizontal means at rest; straight sloping means uniform motion; a curve means non-uniform; steeper means faster.
- A line from the origin to gives ; from to gives , not .
- Three stages , at rest, over s give an average of .
- A distance-time graph can never fall or be vertical — a falling one must be displacement-time.
- Velocity-time graph: the slope is the acceleration and the region under it is the distance.
- The units prove it: .
- From to : and m, matching .
- Horizontal at for s: , m.
- Falling from to rest in s: (a retardation of ) and m.
- Three-phase bus: m in s, so about average.
- Area below the velocity axis is negative displacement, so the region gives displacement in general and distance only without reversal.
- Sketches: at rest — both horizontal, the velocity graph on the axis; uniform velocity — straight sloping distance graph, horizontal velocity graph; uniform acceleration — curved distance graph, straight rising velocity graph; uniform retardation — flattening curve, straight falling line.
- From rest at : gives m, and the triangle at has area m.
- Uniform acceleration gives a straight velocity graph and a curved distance graph — never the same shape twice.
- Graphical derivations from a straight line to :
- slope gives , so ****;
- area gives , and substituting gives ****;
- substituting gives , so **.
- The trapezium splits into a rectangle and a triangle ** — the starting speed's contribution and the acceleration's.
- Checked at , , : , m, .
- All three equations need UNIFORM acceleration — the derivation used a straight line.
- Distances in successive seconds from rest go as ****, since sum to .
Sketch a velocity-time graph for your bus ride to school from memory, then work out the distance from the area and compare it with the real journey length.
- A line from the origin to gives ; from to gives , not .
- Three stages , at rest, over s give an average of .
- A distance-time graph can never fall or be vertical — a falling one must be displacement-time.
- Velocity-time graph: the slope is the acceleration and the region under it is the distance.
- The units prove it: .
- From to : and m, matching .
- Horizontal at for s: , m.
- Falling from to rest in s: (a retardation of ) and m.
- Three-phase bus: m in s, so about average.
- Area below the velocity axis is negative displacement, so the region gives displacement in general and distance only without reversal.
- Sketches: at rest — both horizontal, the velocity graph on the axis; uniform velocity — straight sloping distance graph, horizontal velocity graph; uniform acceleration — curved distance graph, straight rising velocity graph; uniform retardation — flattening curve, straight falling line.
- From rest at : gives m, and the triangle at has area m.
- Uniform acceleration gives a straight velocity graph and a curved distance graph — never the same shape twice.
- Graphical derivations from a straight line to :
- slope gives , so ****;
- area gives , and substituting gives ****;
- substituting gives , so **.
- The trapezium splits into a rectangle and a triangle ** — the starting speed's contribution and the acceleration's.
- Checked at , , : , m, .
- All three equations need UNIFORM acceleration — the derivation used a straight line.
- Distances in successive seconds from rest go as ****, since sum to .
Sketch a velocity-time graph for your bus ride to school from memory, then work out the distance from the area and compare it with the real journey length.