Velocity and Types of Velocity concepts and solved Numerical
Velocity and Types of Velocity concepts and solved Numerical
Velocity :
Definition : Velocity is the rate of change of displacement of a body with respect to time.
Tells us : How fast and in which direction an object is moving.
Quantity : It is a Vector Quantity It means, it gives both Magnitude (Speed) as well as direction.
SI unit : SI unit of Velocity is meter per second (m/s)
Formula : v = Δs / Δt
where,
v
= velocity
Δs
= displacement
Δt
= time taken
Difference between Speed and Velocity :
|
Aspect |
Speed |
Velocity |
|
Definition |
Distance travelled per unit time. |
Displacement per unit time. |
|
Quantity type |
Scalar quantity (only magnitude). |
Vector quantity (magnitude + direction). |
|
Formula |
Speed = Distance ÷ Time |
Velocity = Displacement ÷ Time |
|
Value |
Always positive (or zero). |
Can be positive, negative, or zero (depending on
direction). |
|
Path considered |
Depends on total path length (distance). |
Depends on shortest path (displacement). |
|
Example |
A car moves at 60 km/h. |
A car moves at 60 km/h east. |
|
Average
calculation |
Average Speed = Total Distance ÷ Total Time. |
Average Velocity = Net Displacement ÷ Total Time. |
Types of Velocity :
1) Uniform Velocity :
Definition: A body has uniform velocity when its
magnitude and direction remain constant with time.
- Equivalently, it undergoes equal
displacements in equal time intervals in a straight line.
Implications:
- Acceleration a = 0 (no
change in speed or direction)
- Motion is rectilinear
- Average velocity = instantaneous velocity = constant vector v
Core Formulae
1. Constant
velocity: v = Δs / Δt = constant
2. Displacement–time law: s(t) = so + v t
3. Position–time law (1D): x(t) = xo + v t
4. Average velocity: vavg = [s(t2) – s(t1) / (t2 – t1)] = v
5. Distance
covered: s = v t
6. Acceleration: a = dv/dt = 0
Graphs of Uniform Velocity
1) Displacement–time (s–t) graph: Straight line with slope = velocity.
3) Acceleration–time (a–t) graph: Line on the time axis (a = 0).
Solved Numerical:
1) Displacement from speed and time
Question:
A scooter moves with uniform speed v = 15 m/s for t = 12 s. Find displacement.
Solution:
s = v × t = 15 × 12 = 180 m.
Answer: 180 m.
2) Position after a given time
Question:
A particle starts at x₀ = -20 m with uniform velocity v = 5 m/s. Find x at t = 18 s.
Solution:
x = x₀ + v t = -20 + 5× 18 = 70 m.
Answer: 70 m.3) Time to cover a distance
Question:
A car moves uniformly at 72 km/h. How long to cover 1.2 km?
Solution:
v = 20 m/s, s = 1200 m, t = s/v = 1200/20 = 60
s.
Answer: 60 s.
4) Relative velocity
Question:
Cyclist A = 5 m/s east, B = 3 m/s east. Find velocity of A relative to B and
separation after 40 s.
Solution:
VA/B
= 2 m/s east;
Separation
= 2 × 40 = 80 m.
Answer:
2 m/s east, 80 m.
5) Displacement from v–t graph
Question:
A body moves with v = 12 m/s for 2 min. Find displacement.
Solution:
t = 120 s;
s
= v t = 12 × 120 = 1440 m.
Answer:
1440 m.
6) Uniform velocity in 2D
Question:
v = 6i - 8j m/s. Find displacement in 10 s, its magnitude and direction.
Solution:
Δs
= 60i - 80j
|Δs| = 100 m, θ = -53.13° below +x.
Answer:
100 m at 53.13° below +x.
7) Reading velocity from s–t graph
Question:
An s–t line slope = 4 m/s. Find displacement in 25 s.
Solution:
s = v t = 4 × 25 = 100 m.
Answer:
100 m.
8) Head-on approach
Question:
A at 0 m with v = 10 m/s right, B at 200 m with v = 5 m/s left. When and where
meet?
Solution:
10t
= 200 - 5t
t = 13.33 s, x = 133.33 m.
Answer:
Meet at 13.33 s at 133.33 m.
Practice Questions
1) Train at 54 km/h
for 5 min: displacement? (Answer: 4500 m)
2) Object at x₀ = 25 m, v = -2 m/s. Position at t = 9 s? (Answer:
7 m)
3) v = 3i + 4j m/s. Displacement in 15 s?
Magnitude? (Answer: 75 m)
2) Non -uniform Velocity :
Definition : A body is said to have non-uniform
velocity when either its magnitude (speed) or direction or both change with
time.
·
The body covers unequal displacements in equal intervals of
time, or it does not move in a straight line at constant speed.
Examples:
· A car speeding up or slowing down.
· A body moving in a circular path at constant speed (direction changes)
Core Formulae
1. Instantaneous
velocity: v = ds/dt
2. Average
velocity: vavg = Δs / Δt = (s₂ - s₁)/(t₂ - t₁)
3. Uniform
acceleration (special case):
v = u + at
s = ut + ½at²
v² - u² = 2as
4. General case: s = ∫v(t) dt, v(t) = ds/dt
Graphical Representation
1) Displacement–time (s–t) graph: Curve (not a straight line). Slope at a point = instantaneous velocity.
2) Velocity–time (v–t) graph: Not horizontal; slope = acceleration, area = displacement.
3) Acceleration–time (a–t) graph: May vary; line above/below axis depending on acceleration.
Solved Numerical:
1) Accelerated motion (straight line)
Question:
A car starts from rest and accelerates uniformly at a = 2 m/s² for t = 5 s.
Find final velocity and displacement.
Solution:
u
= 0
v = u + at = 10 m/s
s
= ut + ½at² = 25 m.
2) Retardation (slowing down)
Question:
A bike moving at u = 20 m/s comes to rest in t = 5 s. Find retardation and
distance covered.
Solution:
a = (v - u)/t
= -4
m/s²
s = ut + ½at²
s = 50
m.
3) Average velocity with varying speeds
Question:
A bus travels first 60 km at 40 km/h, and next 60 km at 60 km/h. Find average
velocity.
Solution:
Time = 1.5 h + 1 h = 2.5 h
Vavg = 120/2.5
Vavg = 48 km/h.
4) Non-linear displacement relation
Question:
A particle moves such that s = 4t² (m). Find instantaneous velocity at t = 3 s.
Solution:
v = ds/dt
= 8t
at
t = 3, v = 24 m/s.
5) Circular motion
Questuon:
A body moves in a circle of radius 2 m with constant speed 6 m/s. Why is
velocity non-uniform?
Solution:
Magnitude constant (6 m/s), direction changes
continuously. Hence non-uniform.
Key Points / Pitfalls
- Speed constant
but direction changing → still non-uniform velocity (e.g., circular motion).
- Average velocity is not equal to instantaneous velocity.
- For variable velocity, always use calculus or graph interpretation
(slope/area).
3) Average Velocity :-
Definition: Average velocity is defined as the total displacement of a body divided by the total time taken.
- It tells us the overall rate of change of position of an object during a given time interval.
Important Points
1. Average velocity depends on displacement, not on distance.
2. If the motion is along a straight line in the same direction, average velocity = average speed.
3. If the body returns to the starting point, displacement = 0, hence average velocity = 0.
4. It is a vector quantity, having both magnitude and direction.
Formula:
1) Average Velocity (vavg) = Total Displacement / Total Time
Mathematically:
vavg
= Δs / Δt
2) Average Velocity Formula (When Equal Distance
is Covered)
When equal distances are covered at two different speeds, the average velocity (or average speed) is NOT the arithmetic mean but the harmonic mean of the two speeds.
Mathematically:
where:
v1
= speed during first part of the journey
v2
= speed during second part of the journey
Derivation:
1. Let the distance
covered in each part = d
2. Time taken in first part = d / v1
3. Time taken in second part = d / v2
4. Total distance = 2d
5. Total time = (d / v1) + (d / v2)
Therefore,
vavg =
Total Distance / Total Time
= 2d / ((d / v1) + (d / v2))
3) Average Velocity Formula (When Equal Time is
Taken)
When equal time intervals are spent at two different speeds, the average velocity (or average speed) is the arithmetic mean of the two speeds.
Mathematically,
Vavg =
(v1 + v2) / 2
where:
v1 =
speed during first time interval
v2 =
speed during second time interval
Derivation:
1. Let the time in
each part = t
2. Distance covered in first part = v1 × t
3. Distance covered in second part = v2 × t
4. Total distance = (v1 × t) + (v2 × t) = (v1
+ v2) × t
5. Total time = 2t
Therefore,
vavg = Total Distance / Total Time
= ((v1 + v2)
× t) / (2t)
Solved Numerical :
Numerical 1:
A car covers a
displacement of 100 m in 20 s. Find the average velocity.
Solution:
vavg = Δs / Δt = 100 / 20 = 5 m/s
Numerical 2:
A bus moves 500 m
east in 100 s and then 300 m west in 60 s. Find the average velocity.
Solution:
Net displacement = 500 – 300 = 200 m
east
Total time = 100 + 60 = 160 s
vavg =
200 / 160 = 1.25 m/s east
Numerical 3:
A person goes 1200
m north in 600 s and returns to the starting point in the next 600 s. Find the
average velocity for the whole journey.
Solution:
Net displacement = 0 (since final
position = initial position)
Total time = 600 + 600 = 1200 s
vavg= 0
/ 1200 = 0 m/s
Numerical 4:
A train travels at
40 km/h for 1 hour and then at 60 km/h for the next 1 hour. Find the average
velocity of the train.
Solution:
Displacement in 1st hour = 40 km
Displacement in 2nd hour = 60 km
Net displacement = 100 km, Total time = 2 h
vavg =
100 / 2 = 50 km/h
Numerical 5:
A body moves 600 m
north in 2 minutes and then 800 m east in 3 minutes. Find the magnitude of
average velocity.
Solution:
Net displacement = √(600² + 800²) = 1000
m
Total time = 2 + 3 = 5 min = 300 s
vavg =
1000 / 300 ≈ 3.33 m/s (towards northeast)
Numerical 6:
A car travels equal distances with speeds 40 km/h and 60 km/h.
Solution:
vavg = (2 × 40 × 60) / (40 + 60)
= 4800 / 100
= 48 km/h
So, the average speed of the car is 48 km/h.
Numerical 7:
A train moves with
a speed of 40 km/h for 1 hour and then 60 km/h for the next 1 hour.
vavg = (40 + 60) / 2
= 100 / 2
= 50 km/h
So, the average speed of the train is 50 km/h.
4) Instantaneous Velocity :-
Instantaneous Velocity
The concept of
instantaneous velocity is very important in physics. It helps us to describe
the motion of an object at a particular moment of time, instead of over a long
interval of time.
Definition:
- Instantaneous velocity is the velocity of an object at a particular instant of time.
- It is the rate of change of displacement with respect to time at that specific moment.
Formula:
Mathematically,
instantaneous velocity is expressed using calculus:
v = lim (Δt → 0)
(Δs / Δt)
v = ds/dt
where:
v = instantaneous
velocity
s = displacement
t = time
Explanation with Graph:
On a displacement–time
(s–t) graph:
- Average velocity between two points is the slope of the secant line.
- Instantaneous velocity at a point is the slope of the tangent to the curve at that point.
Key Points:
1. Instantaneous
velocity can be positive, negative, or zero.
2. It is a vector
quantity (it has both magnitude and direction).
3. If the velocity
is constant, then instantaneous velocity = average velocity.
4. In non-uniform
motion, instantaneous velocity keeps changing with time.
Solved Numerical:
Example 1:
The displacement of
a particle moving in a straight line is given by s = 5t² + 2t (in meters),
where t is in seconds. Find the instantaneous velocity at t = 3 s.
Solution:
s =
5t² + 2t
v =
ds/dt
=
d(5t² + 2t)/dt
v
= 10t + 2
At t = 3 s: v = 10(3) + 2 = 32 m/s
Therefore, the instantaneous velocity at 3 seconds is 32 m/s.
Example 2:
The position of a
car is given by s = 4t³ (in meters). Find its instantaneous velocity at t = 2
s.
Solution:
s =
4t³
v =
ds/dt
= d(4t³)/dt
v = 12t²
At, t = 2 s:
v = 12(2²)
v = 48 m/s
Therefore, the instantaneous velocity of the car at 2 seconds is 48 m/s.
Example 3:
The displacement of
a body is given by s = 50 + 20t - 5t². Find its instantaneous velocity at t = 4
s.
Solution:
s = 50
+ 20t - 5t²
v =
ds/dt
v = 20 - 10t
At, t = 4 s:
v = 20 -
10(4)
v = -20 m/s
The negative sign indicates that the body is moving in the opposite direction.
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