PhysicsCBSE Class 11 · Motion in a Straight Line

Kinematic Equations and Motion Graphs Class 11

Updated 17 October 2026 · Fact-checked

For constant acceleration a along a line: v = u + at, s = ut + ½at² and v² = u² + 2as, and the distance in the nth second is u + a(2n − 1)/2. On an x-t graph the slope is velocity; on a v-t graph the slope is acceleration and the area under the curve is displacement. For variable acceleration use v = dx/dt, a = dv/dt and integrate.

Slope of the v-t line is acceleration; the area under it is displacement, the rectangle ut plus the triangle ½at², so s = ut + ½at².

Illustration drawn for QuestionBuddy.

Understand Kinematic Equations and Motion Graphs

Equations for uniform acceleration. If a particle moves along a straight line with constant acceleration $a$ , starting with velocity $u$ :
$ $v = u + a t, s = u t + \frac{1}{2} a t^{2}, v^{2} = u^{2} + 2 a s, s = \frac{( u + v )}{2} t .$ $
The distance covered in the $n$ th second only is $s_{n} = u + \frac{a}{2} (2 n - 1)$ , obtained as $s (n) - s (n - 1)$ . The equations hold for signed quantities, so choose a positive direction and keep signs.

Graphs.

*x-t graph:* slope $=$ velocity. A straight line means uniform velocity; a curve means acceleration; a horizontal line means rest.
*v-t graph:* slope $=$ acceleration; **area between the curve and the t-axis $=$ displacement**. Areas above the axis count positive, below negative. For total *distance*, add the magnitudes of the areas.
*a-t graph:* area gives the change in velocity.

Calculus form. For variable acceleration, $v = \frac{d x}{d t}$ , $a = \frac{d v}{d t} = v \frac{d v}{d x}$ . Integrate: $v = u + \int a d t$ and $x = x_{0} + \int v d t$ . For example if $x = 2 t^{3} - 9 t^{2} + 12 t + 1$ then $v = 6 t^{2} - 18 t + 12$ and $a = 12 t - 18$ .

Average speed versus average velocity. Average speed is total distance over total time; average velocity is displacement over time. If a body covers two equal distances at speeds $v_{1}$ and $v_{2}$ , the average speed is the harmonic mean $\frac{2 v _{1} v _{2}}{v _{1} + v _{2}}$ , not the arithmetic mean.

Chase problems. Write the position of each body as a function of time with a common origin and clock, equate them to find the meeting time, and check that the time is positive. The gap is greatest when the relative velocity is zero.

Kinematic Equations and Motion Graphs: formulas and key facts

First equation: $v = u + a t$; No s.
Second equation: $s = u t + \frac{1}{2} a t^{2}$; No v.
Third equation: $v^{2} = u^{2} + 2 a s$; No t.
Distance in the nth second: $s_{n} = u + \frac{a}{2} (2 n - 1)$; Uniform acceleration.
Calculus relations: $v = \frac{d x}{d t}, a = \frac{d v}{d t} = v \frac{d v}{d x}$; Variable acceleration.
Mean speed over equal distances: $\overset{v}{ˉ} = \frac{2 v _{1} v _{2}}{v _{1} + v _{2}}$; Harmonic mean.

How to solve Kinematic Equations and Motion Graphs questions

Decide first whether acceleration is constant, then pick the equation that leaves out the unknown you do not need.

1Choose the positive direction and list u, v, a, s, t with signs.
2If a is constant, choose the equation that omits the quantity neither given nor asked.
3For a graph, read slopes for velocity or acceleration and areas for displacement; split into triangles and rectangles.
4If x(t) is given, differentiate for v and a; set v = 0 for turning points.
5For total distance, add the lengths of each stretch between turning points.
6For chases, equate positions and test that the answer is physical.

Kinematic Equations and Motion Graphs in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A bus pulling out of a stop

Suppose a bus starts from rest with acceleration 2 m/s² for 9 s, then runs at constant speed for 20 s, then brakes uniformly to rest in 12 s.

Find the maximum speed and the total distance.

Try it first, then show the solutionSolution

v = 0 + 2 × 9 = 18 m/s.
Distances: ½ × 2 × 81 = 81 m; 18 × 20 = 360 m; ½ × 18 × 12 = 108 m.
Total = 81 + 360 + 108 = 549 m.

Answer: 18 m/s and 549 m.

Memory hook: Area of the trapezoid under the v-t graph is the distance.

Merging onto a highway

Suppose a slow truck cruises at 16 m/s and a car 40 m behind it at 16 m/s begins to accelerate at 0.5 m/s² to overtake.

When does the car draw level?

Try it first, then show the solutionSolution

Car position (from its start): 16t + 0.25t². Truck: 40 + 16t.
Equate: 0.25t² = 40, t² = 160, t = 12.6 s.

Answer: About 12.6 s.

Memory hook: Equal speeds first, so the whole gap is made up by the extra acceleration term.

Tricks to remember Kinematic Equations and Motion Graphs

Short memory aids. Say each one aloud once, then use it on a question the same day.

SUVAT missing-letter rule
Each equation misses one letter: v = u + at (no s), s = ut + ½at² (no v), v² = u² + 2as (no t).
Use it when: Selecting the equation.
Slope and area
x-t slope gives v; v-t slope gives a; v-t area gives displacement.
Use it when: Graph questions.
1:3:5
From rest, distances in successive equal seconds are in the ratio 1 : 3 : 5 : 7 …
Use it when: Quick estimates.
Turning point
v = 0 means the particle reverses; distance then splits at that time.
Use it when: Distance versus displacement.

Kinematic Equations and Motion Graphs: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Uniform acceleration · 2 marks · Very short answerBoard level

A cyclist's speed rises uniformly from 4 m/s to 16 m/s in 6 s. Find her acceleration and the distance covered in this time.

Try it first, then show the solutionSolution

a = (16 − 4)/6 = 2 m/s². (1)
s = (u + v)t/2 = (4 + 16)/2 × 6 = 60 m. (1)

Answer: a = 2 m/s²; s = 60 m.

Example 2: Reading a v-t graph · 3 marks · Short answerBoard level

A v-t graph is a trapezoid: velocity rises uniformly from 0 to 20 m/s in 5 s, stays 20 m/s for 10 s, then falls uniformly to 0 in 5 s. Find the total distance and the average speed.

Try it first, then show the solutionSolution

Total time = 20 s; the shape is a trapezoid with parallel sides 20 s and 10 s and height 20 m/s. (1)
Distance = ½(20 + 10) × 20 = 300 m. (1)
Average speed = 300/20 = 15 m/s. (1)

Answer: 300 m; 15 m/s.

Example 3: Distance in the nth second · 3 marks · Short answerHarder board level

A body moving along a straight line with uniform acceleration 4 m/s² has initial speed 5 m/s. Find the distance it covers in the 6th second.

Try it first, then show the solutionSolution

s₆ = u + a(2n − 1)/2 with n = 6. (1)
= 5 + 4 × 11/2 = 5 + 22 = 27 m. (1)
Check: s(6) − s(5) = (30 + 72) − (25 + 50) = 102 − 75 = 27 m. (1)

Answer: 27 m.

Example 4: Catching a speeding car · 4 marks · Long answerHarder board level

A speeder passes a stationary police car at a constant 20 m/s. The police car starts from rest at that instant with a = 2 m/s². Find (a) the time and place of catching up, (b) the police car's speed then, (c) the greatest separation.

Try it first, then show the solutionSolution

(a) 20t = ½ × 2 × t² = t², so t = 20 s; distance = 20 × 20 = 400 m. (1)
(b) v = 2 × 20 = 40 m/s. (1)
(c) The gap is greatest when speeds are equal: 2t = 20, t = 10 s. (1)
Gap = 20 × 10 − 10² = 100 m. (1)

Answer: (a) 20 s, 400 m; (b) 40 m/s; (c) 100 m.

Example 5: A particle with polynomial position · 4 marks · Long answerHOTS / competency

A particle moves along x with x = 2t³ − 9t² + 12t + 1 (x in m, t in s). Find (a) the times at which it is momentarily at rest, (b) its acceleration at those times, (c) the distance travelled in the first 3 s.

Try it first, then show the solutionSolution

v = dx/dt = 6t² − 18t + 12 = 6(t − 1)(t − 2) = 0 at t = 1 s and 2 s. (1)
a = 12t − 18: a(1) = −6 m/s², a(2) = +6 m/s². (1)
x(0) = 1, x(1) = 6, x(2) = 5, x(3) = 10 m. (1)
Path: +5 (0 to 1 s), −1 (1 to 2 s), +5 (2 to 3 s); distance = 11 m (displacement only 9 m). (1)

Answer: (a) 1 s and 2 s; (b) −6 and +6 m/s²; (c) 11 m.

Where students lose marks in Kinematic Equations and Motion Graphs

The slips examiners see most often, and what to write instead.

Mistake: Using the average of u and v when acceleration is not constant.
Applying the shortcut where it does not hold.
Do this: Use (u + v)/2 only for uniform acceleration.
Usually costs: The method marks.
Mistake: Treating displacement as distance in a graph with a negative part.
Adding signed areas.
Do this: Add magnitudes of areas for distance.
Usually costs: 1 mark.
Mistake: Averaging speeds arithmetically for equal distances.
Taking (v₁ + v₂)/2.
Do this: Use total distance over total time (harmonic mean).
Usually costs: The answer mark.
Mistake: Not checking the sign of a root in chase problems.
Accepting t = 0 or negative times.
Do this: Reject unphysical roots.
Usually costs: 1 mark.

Exam tips

State which equation you use and why (the missing quantity).
Draw the v-t graph first for multi-stage motion; areas are faster than three equations.
For x(t) questions find the zero-velocity times before the distance.
Give the sign and the unit of every acceleration.
In chase problems write both position equations on the same origin.

Practice questions on Kinematic Equations and Motion Graphs

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The area between an acceleration-time graph and the time axis, from t₁ to t₂, gives the
(a) displacement (b) change in velocity (c) change in acceleration (d) average speed
Hint
a dt is dv.
Answer
(b) change in velocity.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The area under a v-t graph that goes below the time axis gives distance when counted with its sign.
Reason (R): Areas below the axis represent displacement in the negative direction.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is false but R is true.
(d) Both A and R are false.
Hint
Signed area gives displacement, not distance.
Answer
(c). Signed area gives displacement; distance needs the magnitudes of the areas.
Q3 · 2 marks · Very short answerHarder board level
A car travels the first half of a journey at 40 km/h and the second half at 60 km/h. Find the average speed for the whole journey.
Hint
Equal distances, not equal times.
Answer
Average speed = 2 × 40 × 60/(40 + 60) = 48 km/h.
Q4 · 3 marks · Short answerHarder board level
The velocity of a particle is v = 2 + t³ (SI) and x = 0 at t = 0. Find the acceleration and position at t = 2 s.
Hint
Differentiate for a, integrate for x.
Answer
a = 3t² = 12 m/s²; x = 2t + t⁴/4 = 4 + 4 = 8 m at t = 2 s.
Q5 · 4 marks · Long answerHOTS / competency
A particle has velocity 10 m/s at t = 4 s; its velocity then falls uniformly to −10 m/s at t = 8 s, and it moved with velocity rising uniformly from 0 to 10 m/s during the first 4 s. Find the displacement, the distance and the average velocity for 0 to 8 s.
Hint
The velocity changes sign at t = 6 s.
Answer
0 to 4 s: ½ × 4 × 10 = 20 m. 4 to 6 s: ½ × 2 × 10 = 10 m. 6 to 8 s: −10 m. Displacement = 20 + 10 − 10 = 20 m; distance = 40 m; average velocity = 20/8 = 2.5 m/s; average speed = 5 m/s.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Distance in nth second?Show answer
u + a(2n − 1)/2.
2.Slope of the x-t graph?Show answer
Velocity.
3.Area under a-t graph?Show answer
Change in velocity.
4.Equation without time?Show answer
v² = u² + 2as.
5.a in terms of v and x?Show answer
a = v dv/dx.
6.Ratio of distances in successive seconds from rest?Show answer
1 : 3 : 5 : 7 ...

Kinematic Equations and Motion Graphs: frequently asked questions

Can I use the equations of motion if a changes?

No, unless it is constant over the stretch. Split the motion into stages or use calculus.

Why is the area under a v-t graph the displacement?

Because displacement is the integral of v with respect to time.

What does a negative slope on an x-t graph mean?

The velocity is negative, so the body moves in the negative direction.

Why do speeds not average to give the mean speed?

Average speed weights each speed by the time spent, and for equal distances the slower speed lasts longer.

How do I know a particle has reversed?

Its velocity becomes zero and changes sign.