Three Equations, and the Trick Is Knowing Which One to Pick
Learn the three equations of motion for constant acceleration, how to choose the right one from what a question gives you, how motion in a plane differs from a straight line, and why circular motion is accelerated.
Which equation of motion should you use?
The one that does not contain the quantity you were never given.
There are three equations for constant acceleration, and between them they involve five quantities: initial velocity , final velocity , acceleration , time and displacement . Each equation leaves one of them out:
- has no
- has no
- has no
So list what the question gives you and what it asks for. That will be four of the five quantities, and the fifth — the one nobody mentioned — tells you which equation to use.
A question giving , and and asking for never mentions time, so the third equation is the one. No trial and error is needed. This page covers the third part of the CBSE Class 9 Science chapter on describing motion.
There are three equations for constant acceleration, and between them they involve five quantities: initial velocity , final velocity , acceleration , time and displacement . Each equation leaves one of them out:
- has no
- has no
- has no
So list what the question gives you and what it asks for. That will be four of the five quantities, and the fifth — the one nobody mentioned — tells you which equation to use.
A question giving , and and asking for never mentions time, so the third equation is the one. No trial and error is needed. This page covers the third part of the CBSE Class 9 Science chapter on describing motion.
Formula
What are the three equations of motion?
For an object moving with constant acceleration:
A fourth relation is often useful, and follows from the first two:
Here is initial velocity, final velocity, acceleration, time and displacement, all in SI units.
Where they come from. The first is just the definition of acceleration rearranged, since . The fourth says displacement equals average velocity times time, and for constant acceleration the average velocity is — which is the area of the trapezium under the velocity-time graph from the previous part of this chapter.
Worked example 1 — the straightforward case. A body starts from rest and accelerates at for s.
Cross-check with the third equation: , so m/s. The two agree.
Worked example 2 — braking. A car travelling at m/s brakes with a retardation of . Find the time to stop and the distance covered.
Here because the acceleration opposes the motion, and at the end.
Check with the second equation: m. Correct.
Worked example 3 — three unknowns from three givens. A body starts at m/s, accelerates at , and covers m. Find its final velocity and the time taken.
Time is not given and is wanted later, so start with the equation that omits it:
Now the first equation gives the time:
Check with the second equation: m. Correct.
The sign convention decides everything. Choose a positive direction, then keep it for the whole problem. A retardation gets a minus sign; it is not a separate quantity to be substituted as positive. Worked example 2 would have given a negative stopping distance if the sign had been dropped.
All three need constant acceleration. They do not apply to non-uniform acceleration, and they do not apply to circular motion at changing speed — a limit taken up later on this page.
A fourth relation is often useful, and follows from the first two:
Here is initial velocity, final velocity, acceleration, time and displacement, all in SI units.
Where they come from. The first is just the definition of acceleration rearranged, since . The fourth says displacement equals average velocity times time, and for constant acceleration the average velocity is — which is the area of the trapezium under the velocity-time graph from the previous part of this chapter.
Worked example 1 — the straightforward case. A body starts from rest and accelerates at for s.
Cross-check with the third equation: , so m/s. The two agree.
Worked example 2 — braking. A car travelling at m/s brakes with a retardation of . Find the time to stop and the distance covered.
Here because the acceleration opposes the motion, and at the end.
Check with the second equation: m. Correct.
Worked example 3 — three unknowns from three givens. A body starts at m/s, accelerates at , and covers m. Find its final velocity and the time taken.
Time is not given and is wanted later, so start with the equation that omits it:
Now the first equation gives the time:
Check with the second equation: m. Correct.
The sign convention decides everything. Choose a positive direction, then keep it for the whole problem. A retardation gets a minus sign; it is not a separate quantity to be substituted as positive. Worked example 2 would have given a negative stopping distance if the sign had been dropped.
All three need constant acceleration. They do not apply to non-uniform acceleration, and they do not apply to circular motion at changing speed — a limit taken up later on this page.
How do you apply the equations to a falling object?
**Replace by , the acceleration due to gravity**, and keep the sign convention consistent.
Near the Earth's surface a freely falling body accelerates downward at , which can be taken as for calculation unless a question gives . The mass of the object does not appear anywhere in the equations, so a heavy stone and a light one fall together when air resistance can be ignored.
Worked example 1 — dropped from a height. A stone is dropped from a tower m tall. Find the time to reach the ground and its speed on landing.
Taking downward as positive, , , m:
Check with the third equation: , so m/s. Correct.
Worked example 2 — thrown upward. A ball is thrown straight up at m/s. Find the time to reach the top and the maximum height.
Taking upward as positive, gravity now opposes the motion, so . At the highest point the ball is momentarily at rest, so :
The ball is at rest at the top, and still accelerating. Its velocity is zero for an instant while gravity continues to act, so throughout the flight — on the way up, at the top and on the way down. Zero velocity does not mean zero acceleration, and this is the standard example used to prove it.
Worked example 3 — the whole flight. The same ball returns to the thrower's hand. By symmetry it takes another s to come down and arrives at m/s downward, so the total flight lasts s. Its displacement over the whole flight is zero, while the distance covered is m — the distinction from the first part of this chapter, appearing again.
Everyday evidence. A coin and a sheet of paper dropped together land at different times, which seems to contradict all of this. Crumple the paper into a tight ball and they land together. Nothing about gravity changed; air resistance, which the equations ignore, mattered for the flat sheet and not for the ball.
Pick the positive direction and write it down. Taking downward as positive makes a falling-body problem free of minus signs; taking upward as positive makes a thrown-ball problem read naturally. Either is correct — mixing them inside one problem is not.
Near the Earth's surface a freely falling body accelerates downward at , which can be taken as for calculation unless a question gives . The mass of the object does not appear anywhere in the equations, so a heavy stone and a light one fall together when air resistance can be ignored.
Worked example 1 — dropped from a height. A stone is dropped from a tower m tall. Find the time to reach the ground and its speed on landing.
Taking downward as positive, , , m:
Check with the third equation: , so m/s. Correct.
Worked example 2 — thrown upward. A ball is thrown straight up at m/s. Find the time to reach the top and the maximum height.
Taking upward as positive, gravity now opposes the motion, so . At the highest point the ball is momentarily at rest, so :
The ball is at rest at the top, and still accelerating. Its velocity is zero for an instant while gravity continues to act, so throughout the flight — on the way up, at the top and on the way down. Zero velocity does not mean zero acceleration, and this is the standard example used to prove it.
Worked example 3 — the whole flight. The same ball returns to the thrower's hand. By symmetry it takes another s to come down and arrives at m/s downward, so the total flight lasts s. Its displacement over the whole flight is zero, while the distance covered is m — the distinction from the first part of this chapter, appearing again.
Everyday evidence. A coin and a sheet of paper dropped together land at different times, which seems to contradict all of this. Crumple the paper into a tight ball and they land together. Nothing about gravity changed; air resistance, which the equations ignore, mattered for the flat sheet and not for the ball.
Pick the positive direction and write it down. Taking downward as positive makes a falling-body problem free of minus signs; taking upward as positive makes a thrown-ball problem read naturally. Either is correct — mixing them inside one problem is not.
How is motion in a plane different from motion in a straight line?
In a straight line a sign is enough to give direction; in a plane you need two numbers or a direction in words.
Along a single axis there are only two possible directions, so and describe them completely. That is what made every calculation on this page manageable.
In a plane there are infinitely many directions, so a vector needs either its two components — how far east and how far north — or a magnitude together with a stated direction.
Worked example — the same two moves, arranged two ways.
Move m east and then m east. Both are along one line, so they simply add:
Now move m east and then m north. These are at right angles, so they do not add arithmetically:
The distance is m in both cases. The displacement is m in the first and m in the second — which is exactly why direction cannot be ignored.
Velocity in a plane can change without changing in size. A car going round a bend at a steady km/h has a constant speed and a continuously changing velocity, because its direction is changing. In a straight line that could not happen: keeping the speed constant would keep the velocity constant too.
Two familiar examples of plane motion. A ball thrown at an angle moves forward and rises or falls at the same time, so its motion has a horizontal part and a vertical part. A stone whirled on a string moves in a circle and changes direction at every instant.
So a plane needs vectors, not just signs. This is the reason the next section can say something that would be impossible in one dimension: that an object can accelerate while its speed never changes at all.
A common misreading. Motion in a plane does not mean motion in the air or motion of an aeroplane. A plane here is a flat surface — a football rolling across a level ground is moving in a plane, and so is a coin sliding across a table.
Along a single axis there are only two possible directions, so and describe them completely. That is what made every calculation on this page manageable.
In a plane there are infinitely many directions, so a vector needs either its two components — how far east and how far north — or a magnitude together with a stated direction.
Worked example — the same two moves, arranged two ways.
Move m east and then m east. Both are along one line, so they simply add:
Now move m east and then m north. These are at right angles, so they do not add arithmetically:
The distance is m in both cases. The displacement is m in the first and m in the second — which is exactly why direction cannot be ignored.
Velocity in a plane can change without changing in size. A car going round a bend at a steady km/h has a constant speed and a continuously changing velocity, because its direction is changing. In a straight line that could not happen: keeping the speed constant would keep the velocity constant too.
Two familiar examples of plane motion. A ball thrown at an angle moves forward and rises or falls at the same time, so its motion has a horizontal part and a vertical part. A stone whirled on a string moves in a circle and changes direction at every instant.
So a plane needs vectors, not just signs. This is the reason the next section can say something that would be impossible in one dimension: that an object can accelerate while its speed never changes at all.
A common misreading. Motion in a plane does not mean motion in the air or motion of an aeroplane. A plane here is a flat surface — a football rolling across a level ground is moving in a plane, and so is a coin sliding across a table.
Why is uniform circular motion accelerated motion?
Because the direction of the velocity changes continuously, even though its size does not.
An object in uniform circular motion moves along a circular path at constant speed. At every instant its velocity points along the tangent to the circle, and the tangent direction is different at every point. A changing velocity is an acceleration, so circular motion is accelerated motion — always.
The acceleration points towards the centre of the circle. That is why a stone whirled on a string needs the string: something must pull it inward continuously, and when the string breaks the stone flies off along the tangent rather than outward.
The speed in a circular path. In one full revolution the object covers the circumference in the time , called the time period:
Worked example 1. An athlete runs once around a circular track of radius m in s.
Worked example 2. A stone tied to a string of length m is whirled in a horizontal circle, completing one round in s.
Worked example 3. A cyclist rides once around a circular park of diameter m in s.
The radius is m, so the circumference is again m:
Worked example 4 — speed against velocity over one round. For that cyclist, the average speed over one complete round is m/s. The displacement, however, is zero, so
The cyclist's speed was never zero for an instant, and the average velocity is exactly zero — the lap result from the first part of this chapter, arrived at from the circular-motion formula.
Uniform does not mean unaccelerated here. Uniform circular motion means uniform in speed. Its velocity is non-uniform, so it is accelerated — and a question asking whether a body moving at constant speed can be accelerating is asking about precisely this case. The answer is yes, and circular motion is the example.
Halve the diameter before substituting. Worked example 3 gave the diameter, and using m as the radius would have doubled the answer. This is the same reading error that spoils cylinder calculations, and the fix is the same: write down on its own line before using it.
An object in uniform circular motion moves along a circular path at constant speed. At every instant its velocity points along the tangent to the circle, and the tangent direction is different at every point. A changing velocity is an acceleration, so circular motion is accelerated motion — always.
The acceleration points towards the centre of the circle. That is why a stone whirled on a string needs the string: something must pull it inward continuously, and when the string breaks the stone flies off along the tangent rather than outward.
The speed in a circular path. In one full revolution the object covers the circumference in the time , called the time period:
Worked example 1. An athlete runs once around a circular track of radius m in s.
Worked example 2. A stone tied to a string of length m is whirled in a horizontal circle, completing one round in s.
Worked example 3. A cyclist rides once around a circular park of diameter m in s.
The radius is m, so the circumference is again m:
Worked example 4 — speed against velocity over one round. For that cyclist, the average speed over one complete round is m/s. The displacement, however, is zero, so
The cyclist's speed was never zero for an instant, and the average velocity is exactly zero — the lap result from the first part of this chapter, arrived at from the circular-motion formula.
Uniform does not mean unaccelerated here. Uniform circular motion means uniform in speed. Its velocity is non-uniform, so it is accelerated — and a question asking whether a body moving at constant speed can be accelerating is asking about precisely this case. The answer is yes, and circular motion is the example.
Halve the diameter before substituting. Worked example 3 gave the diameter, and using m as the radius would have doubled the answer. This is the same reading error that spoils cylinder calculations, and the fix is the same: write down on its own line before using it.
Exam tip
Exam tip: list your five quantities before choosing an equation
**Write down , , , and , filling in what you know.** The one quantity neither given nor asked for names the equation: no means ; no means ; no means .
Fix a positive direction and keep it. A retardation takes a minus sign — — and is not substituted as positive.
For a dropped object take downward as positive with ; for one thrown up take upward as positive with . Use unless told otherwise.
**At the highest point but is still . Zero velocity never means zero acceleration.
Convert to SI units before substituting** — km/h to m/s by multiplying by .
Check your answer with a second equation. In worked example 3, m/s and s must reproduce m, and they do.
In a plane, perpendicular displacements do not add arithmetically: m east and m north give m, not m.
Uniform circular motion is accelerated — constant speed, changing direction, acceleration towards the centre.
Use , and halve the diameter first when a diameter is given.
And remember all three equations need constant acceleration — they do not apply otherwise.
Fix a positive direction and keep it. A retardation takes a minus sign — — and is not substituted as positive.
For a dropped object take downward as positive with ; for one thrown up take upward as positive with . Use unless told otherwise.
**At the highest point but is still . Zero velocity never means zero acceleration.
Convert to SI units before substituting** — km/h to m/s by multiplying by .
Check your answer with a second equation. In worked example 3, m/s and s must reproduce m, and they do.
In a plane, perpendicular displacements do not add arithmetically: m east and m north give m, not m.
Uniform circular motion is accelerated — constant speed, changing direction, acceleration towards the centre.
Use , and halve the diameter first when a diameter is given.
And remember all three equations need constant acceleration — they do not apply otherwise.
Did you know
Why a heavy stone and a light one land together
Look carefully at the equations on this page and notice what is missing from all of them: mass.
, and contain velocity, acceleration, time and displacement, and nothing about how heavy the moving object is. So when the acceleration is , every object gets the same answer.
Drop a small stone and a large one from the same window and they land together. Drop a coin and a flat sheet of paper and they plainly do not — the paper drifts down slowly, and it looks as though the equations have failed.
They have not. What has happened is that the air is exerting an upward force on the paper large enough to matter, because a flat sheet presents a great deal of surface for very little weight. Crumple the same sheet into a tight ball and drop it beside the coin, and the two land together. The paper did not change its mass; it changed its surface.
That is exactly the neglect air resistance condition attached to every free-fall question — not padding, but a statement of when the calculation applies. For a dense compact object over a few metres the correction is negligible. For a feather, a leaf or a sheet of paper it dominates completely.
The same reasoning explains a parachute. Nothing reduces the parachutist's mass or weakens gravity; the canopy simply presents an enormous surface to the air, and the upward force grows until it balances the weight. From then on the velocity stops changing altogether — the acceleration becomes zero while the fall continues at a steady speed.
So mass does not appear in the equations of motion is a genuine and surprising fact about the world, and the everyday exceptions to it are all exceptions about air, never about gravity.
, and contain velocity, acceleration, time and displacement, and nothing about how heavy the moving object is. So when the acceleration is , every object gets the same answer.
Drop a small stone and a large one from the same window and they land together. Drop a coin and a flat sheet of paper and they plainly do not — the paper drifts down slowly, and it looks as though the equations have failed.
They have not. What has happened is that the air is exerting an upward force on the paper large enough to matter, because a flat sheet presents a great deal of surface for very little weight. Crumple the same sheet into a tight ball and drop it beside the coin, and the two land together. The paper did not change its mass; it changed its surface.
That is exactly the neglect air resistance condition attached to every free-fall question — not padding, but a statement of when the calculation applies. For a dense compact object over a few metres the correction is negligible. For a feather, a leaf or a sheet of paper it dominates completely.
The same reasoning explains a parachute. Nothing reduces the parachutist's mass or weakens gravity; the canopy simply presents an enormous surface to the air, and the upward force grows until it balances the weight. From then on the velocity stops changing altogether — the acceleration becomes zero while the fall continues at a steady speed.
So mass does not appear in the equations of motion is a genuine and surprising fact about the world, and the everyday exceptions to it are all exceptions about air, never about gravity.
Exam relevance
How are the equations of motion tested in JEE Main and NEET?
These three equations are the most heavily reused result in the whole of introductory physics, and this page is where they are first met.
This is the foundation for the Class 11 Physics chapter Motion in a Straight Line, examined in both JEE Main and NEET Physics. That chapter derives the same three equations by calculus, adds the relation for the distance covered in the th second, and extends them to graphs of non-uniform acceleration — but the equations themselves are unchanged.
Where they get reused. Class 11 Motion in a Plane applies them separately to the horizontal and vertical components of projectile motion, which is a standing JEE Main topic; the vertical component is exactly the thrown-ball problem worked above. Laws of Motion combines them with , so that a force question becomes a kinematics question once the acceleration is found. And Work, Energy and Power derives the work-energy theorem from directly — multiply that third equation by and the kinetic energy formula appears.
Circular motion becomes centripetal acceleration in Class 11 Motion in a Plane, and then reappears in Gravitation for orbits and in Class 12 Moving Charges and Magnetism for a charged particle in a magnetic field. The Class 9 point that carries all the way through is that constant speed on a curve is still accelerated motion.
What the questions look like. Numericals dominate, and the skill being tested is nearly always equation selection — a question gives three of the five quantities and asks for a fourth, and choosing the wrong equation introduces an unknown you cannot find. Assertion-reason items favour the two counter-intuitive statements on this page: that a body can have zero velocity and non-zero acceleration, and that uniform circular motion is accelerated. Free-fall problems with a ball thrown up from a height, so that up and down distances differ, are a common step up in difficulty.
How board and competitive emphasis differ. A board paper asks you to derive an equation of motion, perhaps from a velocity-time graph, and then sets a single-step numerical. A competitive paper never asks for the derivation; it sets a two-step numerical where the answer of the first equation feeds the second, or where the sign convention has to be handled carefully. Deriving them from the graph is still worth doing once, because it is the clearest reason why the area under a velocity-time graph is the displacement.
The single trap that costs the most marks. Sign errors. A retardation substituted as positive, or taken as positive while upward is the positive direction, produces an answer that is not merely inaccurate but impossible — a negative stopping distance, or a ball that rises for ever. Fix the positive direction in writing before the first substitution, and then let the signs follow it without exception.
A second trap worth naming. Using the equations where the acceleration is not constant. If a question describes an acceleration that changes, or gives a curved velocity-time graph, these three equations do not apply at all — and recognising that is itself the answer the question wants.
This is the foundation for the Class 11 Physics chapter Motion in a Straight Line, examined in both JEE Main and NEET Physics. That chapter derives the same three equations by calculus, adds the relation for the distance covered in the th second, and extends them to graphs of non-uniform acceleration — but the equations themselves are unchanged.
Where they get reused. Class 11 Motion in a Plane applies them separately to the horizontal and vertical components of projectile motion, which is a standing JEE Main topic; the vertical component is exactly the thrown-ball problem worked above. Laws of Motion combines them with , so that a force question becomes a kinematics question once the acceleration is found. And Work, Energy and Power derives the work-energy theorem from directly — multiply that third equation by and the kinetic energy formula appears.
Circular motion becomes centripetal acceleration in Class 11 Motion in a Plane, and then reappears in Gravitation for orbits and in Class 12 Moving Charges and Magnetism for a charged particle in a magnetic field. The Class 9 point that carries all the way through is that constant speed on a curve is still accelerated motion.
What the questions look like. Numericals dominate, and the skill being tested is nearly always equation selection — a question gives three of the five quantities and asks for a fourth, and choosing the wrong equation introduces an unknown you cannot find. Assertion-reason items favour the two counter-intuitive statements on this page: that a body can have zero velocity and non-zero acceleration, and that uniform circular motion is accelerated. Free-fall problems with a ball thrown up from a height, so that up and down distances differ, are a common step up in difficulty.
How board and competitive emphasis differ. A board paper asks you to derive an equation of motion, perhaps from a velocity-time graph, and then sets a single-step numerical. A competitive paper never asks for the derivation; it sets a two-step numerical where the answer of the first equation feeds the second, or where the sign convention has to be handled carefully. Deriving them from the graph is still worth doing once, because it is the clearest reason why the area under a velocity-time graph is the displacement.
The single trap that costs the most marks. Sign errors. A retardation substituted as positive, or taken as positive while upward is the positive direction, produces an answer that is not merely inaccurate but impossible — a negative stopping distance, or a ball that rises for ever. Fix the positive direction in writing before the first substitution, and then let the signs follow it without exception.
A second trap worth naming. Using the equations where the acceleration is not constant. If a question describes an acceleration that changes, or gives a curved velocity-time graph, these three equations do not apply at all — and recognising that is itself the answer the question wants.
Key takeaways
Equations of motion and circular motion: quick revision
- For constant acceleration: (no ), (no ), (no ), and .
- Choose by elimination: list , and the quantity neither given nor wanted names the equation.
- , , s gives m/s and m, confirmed by .
- Braking: m/s with gives s and m, confirmed two ways.
- m/s, , m gives so m/s, then s — and m checks it.
- Retardation takes a minus sign, and the positive direction must be fixed once for the whole problem.
- Free fall: use , and note that mass does not appear in any equation.
- Dropped from m: gives s and m/s.
- Thrown up at m/s with : s to the top and m. The whole flight lasts s, with distance m and displacement zero.
- **At the top, but still — zero velocity is not zero acceleration.
- A coin and a flat sheet of paper fall differently because of air resistance, not gravity; crumple the paper and they land together.
- In a plane**, direction needs components or words, not just a sign. m east then m east gives m; m east then m north gives m.
- Velocity can change without speed changing — a car rounding a bend at steady speed.
- Uniform circular motion is constant speed with continuously changing direction, so it is accelerated, with the acceleration directed towards the centre.
- . Radius m in s gives m/s; radius m in s gives m/s; diameter m in s gives m/s.
- Over one full round, average speed is non-zero while average velocity is zero.
- Halve the diameter before substituting, and remember all three equations require constant acceleration.
Set yourself one numerical from each equation, solve it, then verify the answer with a different equation — an answer that survives two routes is one you can trust under exam pressure.
- Choose by elimination: list , and the quantity neither given nor wanted names the equation.
- , , s gives m/s and m, confirmed by .
- Braking: m/s with gives s and m, confirmed two ways.
- m/s, , m gives so m/s, then s — and m checks it.
- Retardation takes a minus sign, and the positive direction must be fixed once for the whole problem.
- Free fall: use , and note that mass does not appear in any equation.
- Dropped from m: gives s and m/s.
- Thrown up at m/s with : s to the top and m. The whole flight lasts s, with distance m and displacement zero.
- **At the top, but still — zero velocity is not zero acceleration.
- A coin and a flat sheet of paper fall differently because of air resistance, not gravity; crumple the paper and they land together.
- In a plane**, direction needs components or words, not just a sign. m east then m east gives m; m east then m north gives m.
- Velocity can change without speed changing — a car rounding a bend at steady speed.
- Uniform circular motion is constant speed with continuously changing direction, so it is accelerated, with the acceleration directed towards the centre.
- . Radius m in s gives m/s; radius m in s gives m/s; diameter m in s gives m/s.
- Over one full round, average speed is non-zero while average velocity is zero.
- Halve the diameter before substituting, and remember all three equations require constant acceleration.
Set yourself one numerical from each equation, solve it, then verify the answer with a different equation — an answer that survives two routes is one you can trust under exam pressure.