Equations of Motion

Master the three fundamental equations that describe uniformly accelerated motion! Learn to solve complex motion problems using these powerful mathematical tools.

Three Equations of Motion
When an object moves along a straight line with uniform acceleration, we can relate its velocity, acceleration, and distance using three fundamental equations. These equations are the foundation of kinematics and help us solve complex motion problems.
Equation 1
v = u + at
Velocity-Time Relation
Equation 2
s = ut + ½at²
Position-Time Relation
Equation 3
v² = u² + 2as
Position-Velocity Relation
First Equation of Motion
v = u + at
Where:
v = final velocity (m/s)
u = initial velocity (m/s)
a = acceleration (m/s²)
t = time (s)
Used when time is known
Second Equation of Motion
s = ut + ½at²
Where:
s = displacement (m)
u = initial velocity (m/s)
t = time (s)
a = acceleration (m/s²)
Used when final velocity is unknown
Third Equation of Motion
v² = u² + 2as
Where:
v = final velocity (m/s)
u = initial velocity (m/s)
a = acceleration (m/s²)
s = displacement (m)
Used when time is unknown
Motion Calculator
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Equation Details
v = u + at
This equation relates final velocity to initial velocity, acceleration, and time. It's derived from the definition of acceleration.
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Velocity-Time Graph
From the velocity-time graph, we can derive all three equations:
Slope = acceleration = (v-u)/t
Area = displacement = ut + ½(v-u)t
Graphical Derivation
First Equation (v = u + at):
From the slope of velocity-time graph:
Slope = (v - u) / t = a
Therefore: v = u + at

Second Equation (s = ut + ½at²):
Area under the graph = displacement
Area = Rectangle + Triangle
s = ut + ½(v-u)t
Substituting v = u + at:
s = ut + ½at²

Third Equation (v² = u² + 2as):
From equations 1 and 2, eliminating t:
t = (v-u)/a
s = u(v-u)/a + ½a((v-u)/a)²
Simplifying: v² = u² + 2as
Example 7.5
Problem: A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming uniform acceleration, find (i) the acceleration and (ii) the distance travelled.
Given Data
Given:
u = 0 (starting from rest)
v = 72 km/h = 72 × (5/18) = 20 m/s
t = 5 minutes = 300 s
(i) Finding Acceleration
Using Equation 1:
v = u + at
Rearranging:
a = (v - u) / t
Substituting:
a = (20 - 0) / 300 = 1/15 m/s²
Acceleration = 1/15 m/s² ≈ 0.067 m/s²
(ii) Finding Distance
Using Equation 3:
v² = u² + 2as
Rearranging:
s = (v² - u²) / (2a)
Substituting:
s = (20² - 0²) / (2 × 1/15) = 3000 m = 3 km
Distance travelled = 3 km
Example 7.6
Problem: A car accelerates uniformly from 18 km/h to 36 km/h in 5 s. Calculate (i) the acceleration and (ii) the distance covered.
Given Data
Given:
u = 18 km/h = 18 × (5/18) = 5 m/s
v = 36 km/h = 36 × (5/18) = 10 m/s
t = 5 s
(i) Finding Acceleration
Using Equation 1:
a = (v - u) / t
Substituting values:
a = (10 - 5) / 5 = 1 m/s²
Acceleration = 1 m/s²
(ii) Finding Distance
Using Equation 2:
s = ut + ½at²
Substituting values:
s = 5×5 + ½×1×5² = 25 + 12.5 = 37.5 m
Distance covered = 37.5 m
Braking Example
Problem: Brakes applied to a car produce acceleration of 6 m/s² in opposite direction. If the car takes 2 s to stop, calculate the distance it travels during this time.
Given Data
Given:
a = -6 m/s² (opposite to motion)
t = 2 s
v = 0 m/s (car stops)
Finding Initial Velocity
Using Equation 1:
v = u + at
Rearranging:
u = v - at = 0 - (-6)×2 = 12 m/s
Finding Distance
Using Equation 2:
s = ut + ½at²
Substituting values:
s = 12×2 + ½×(-6)×2² = 24 - 12 = 12 m
Distance travelled = 12 m

Practice Questions

Q1.
A bus starting from rest moves with uniform acceleration of 0.1 m/s² for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Solution:
Given: u = 0, a = 0.1 m/s², t = 2 min = 120 s

(a) Speed acquired:
Using v = u + at
v = 0 + 0.1 × 120 = 12 m/s

(b) Distance travelled:
Using s = ut + ½at²
s = 0×120 + ½×0.1×120² = 720 m
Q2.
A train is travelling at 90 km/h. Brakes are applied to produce uniform acceleration of -0.5 m/s². Find how far the train will go before it stops.
Solution:
Given: u = 90 km/h = 25 m/s, a = -0.5 m/s², v = 0

Using v² = u² + 2as
0² = 25² + 2×(-0.5)×s
0 = 625 - s
s = 625 m

Answer: 625 m
Q3.
A trolley going down an inclined plane has acceleration of 2 cm/s². What will be its velocity 3 s after the start?
Solution:
Given: u = 0 (starts from rest), a = 2 cm/s² = 0.02 m/s², t = 3 s

Using v = u + at
v = 0 + 0.02 × 3 = 0.06 m/s = 6 cm/s

Answer: 6 cm/s
Q4.
A racing car has uniform acceleration of 4 m/s². What distance will it cover in 10 s after start?
Solution:
Given: u = 0 (starts from rest), a = 4 m/s², t = 10 s

Using s = ut + ½at²
s = 0×10 + ½×4×10² = 0 + 2×100 = 200 m

Answer: 200 m
Q5.
A stone is thrown vertically upward with velocity 5 m/s. If acceleration is 10 m/s² downward, find the height attained and time to reach there.
Solution:
Given: u = 5 m/s, a = -10 m/s² (downward), v = 0 (at maximum height)

Height attained:
Using v² = u² + 2as
0² = 5² + 2×(-10)×s
s = 25/20 = 1.25 m

Time to reach:
Using v = u + at
0 = 5 + (-10)×t
t = 0.5 s

Answer: Height = 1.25 m, Time = 0.5 s
Explore Equations of Motion
Equations of Motion
The equations of motion are mathematical relationships that describe the motion of objects under uniform acceleration. These three equations connect velocity, acceleration, time, and displacement, allowing us to solve complex motion problems.
First Equation
v = u + at
Velocity-Time Relation
Second Equation
s = ut + ½at²
Position-Time Relation
Third Equation
v² = u² + 2as
Position-Velocity Relation
When to Use Each Equation
• First Equation (v = u + at): Use when you know or need to find time
• Second Equation (s = ut + ½at²): Use when final velocity is unknown
• Third Equation (v² = u² + 2as): Use when time is unknown

Key Variables:
• u = initial velocity (m/s)
• v = final velocity (m/s)
• a = acceleration (m/s²)
• t = time (s)
• s = displacement (m)
Important Points
• These equations apply only to uniformly accelerated motion
• Acceleration can be positive (speeding up) or negative (slowing down)
• Always convert units to SI units before calculation
• Choose the equation that contains the unknown quantity you need to find
• The equations can be derived graphically from velocity-time graphs
• Sign conventions: Choose positive direction and stick to it throughout