Motion in a Straight Line — Class 11 Physics
1. What is Motion?
An object is in motion when its position changes with time relative to a chosen reference point. Motion is therefore relative: the same object can be at rest with respect to one observer and moving with respect to another.
2. Position and Displacement
For one-dimensional motion, position is represented by a coordinate x along a chosen axis. Displacement is the change in position: Δx = x₂ − x₁. Displacement has direction and can be positive, negative or zero.
Distance is the total path length travelled and is a scalar. Therefore distance is always non-negative and is greater than or equal to the magnitude of displacement.
3. Speed and Velocity
Average speed = total distance / total time. Average velocity = total displacement / total time. Instantaneous velocity is the rate of change of position with time: v = dx/dt.
4. Acceleration
Acceleration is the rate of change of velocity: a = dv/dt. Average acceleration is Δv/Δt. Acceleration may be positive, negative or zero depending on the chosen direction and how velocity changes.
5. Uniformly Accelerated Motion
For constant acceleration, the standard equations are:
- v = u + at
- s = ut + ½at²
- v² = u² + 2as
- s = ((u+v)/2)t
Here u is initial velocity, v final velocity, a acceleration, t time and s displacement. These equations apply to one-dimensional motion with constant acceleration.
6. Graphs of Motion
In a position-time graph, the slope represents velocity. A straight line with constant slope represents constant velocity.
In a velocity-time graph, the slope represents acceleration and the area under the graph represents displacement.
In an acceleration-time graph, the area under the graph represents change in velocity.
Worked Numerical 1 — Average Speed and Velocity
Question: A car travels 60 m east in 10 s and then 40 m west in 10 s. Find its average speed and average velocity.
Solution: Total distance = 60+40=100 m. Total time =20 s. Average speed =100/20=5 m/s.
Taking east as positive, displacement =60−40=20 m east. Average velocity =20/20=1 m/s east.
Worked Numerical 2 — Constant Acceleration
Question: A car starts with velocity 5 m/s and accelerates uniformly at 2 m/s² for 6 s. Find its final velocity and displacement.
Solution: u=5 m/s, a=2 m/s², t=6 s.
v=u+at=5+(2)(6)=17 m/s.
s=ut+½at²=(5)(6)+½(2)(36)=30+36=66 m.
Worked Numerical 3 — Braking
Question: A vehicle moving at 20 m/s comes to rest with uniform acceleration of −4 m/s². Find the stopping distance.
Using v²=u²+2as: 0²=20²+2(−4)s.
8s=400, so s=50 m.
Worked Numerical 4 — Free Fall as One-Dimensional Motion
Question: An object is dropped from rest and falls for 2 s. Take g=9.8 m/s². Find its speed and displacement.
u=0, a=g=9.8 m/s², t=2 s.
v=u+at=19.6 m/s downward.
s=½gt²=½(9.8)(4)=19.6 m downward.
Graph-Based Understanding
If a velocity-time graph is a horizontal line at 10 m/s for 5 s, acceleration is zero and displacement is the rectangular area 10×5=50 m. If velocity changes linearly from 0 to 20 m/s in 4 s, acceleration is 20/4=5 m/s² and displacement is the triangular area ½×4×20=40 m.
Explained MCQs
- An object completes a round trip and returns to its starting point. What is its displacement?
Answer: Zero. Final and initial positions are identical, even though the distance travelled is non-zero. - Which quantity is represented by the slope of a position-time graph?
Answer: Velocity. The slope is change in position divided by change in time. - The area under a velocity-time graph represents:
Answer: Displacement. Velocity multiplied by time gives displacement; for varying velocity the area performs the corresponding integration. - Can an object have zero velocity but non-zero acceleration?
Answer: Yes. At the highest point of a vertically thrown object, instantaneous velocity is zero while acceleration due to gravity remains downward. - If acceleration is negative, does it always mean the object is slowing down?
Answer: No. It depends on the direction of velocity. Negative acceleration can increase speed when velocity is also negative.
Competency Question
A cyclist moves 100 m east and then 100 m west in 40 s. Calculate distance, displacement, average speed and average velocity. Explain why the last two answers are different.
HOTS
A car has positive velocity and negative acceleration. Is it necessarily moving backwards? Explain using the meanings of velocity and acceleration rather than relying only on the signs.
Common Mistakes
- Confusing distance with displacement.
- Confusing average speed with magnitude of average velocity.
- Using the constant-acceleration equations when acceleration is not constant.
- Forgetting to choose a positive direction before assigning signs.
- Reading the slope and area of motion graphs incorrectly.
Quick Revision
- Displacement: Δx=x₂−x₁
- Average velocity: displacement/time
- Average speed: distance/time
- v=dx/dt
- a=dv/dt
- v=u+at
- s=ut+½at²
- v²=u²+2as
- Slope of x-t graph = velocity
- Slope of v-t graph = acceleration
- Area under v-t graph = displacement