Equations of Motion and Graphs ഗ്രാഫ്
Several questions in this chapter need you to find a final velocity, a distance or an acceleration, or to read them from a graph (ഗ്രാഫ്). Examples are the bullet stopping inside a log, the block pushed for 2 s, and Graphs 3.1 and 3.3 in “Let us assess”. This page rebuilds the three equations of motion, shows when to use each one, and teaches you to read distance–time and velocity–time graphs.
- Velocity, acceleration a = (v − u)/t, retardation and km/h → m/s (page 2).
- Squares like 20² = 400, and the area of a rectangle and a triangle.
1. Five quantities, three equations
For motion along a straight line with a constant (uniform) acceleration (സമത്വരണം), five quantities describe everything:
| Symbol | Quantity | Unit |
|---|---|---|
| u | initial velocity (ആദ്യപ്രവേഗം) | m/s |
| v | final velocity (അന്ത്യപ്രവേഗം) | m/s |
| a | acceleration (ത്വരണം), negative for retardation | m/s² |
| t | time taken | s |
| s | displacement (സ്ഥാനാന്തരം), the distance moved along the line | m |
Three equations connect them. Each one leaves out one of the five quantities:
Read the question and list u, v, a, t, s. Mark the three you are given and the one you must find. The fifth one, the quantity that is missing, tells you the equation.
No distance in the question? → v = u + at. No final velocity? → s = ut + ½at². No time? → v² = u² + 2as. Also remember the shortcuts: “starts from rest” means u = 0; “comes to rest” or “stops” means v = 0.
2. v = u + at
This is just the definition of acceleration turned round: a = (v − u)/t ⇒ v − u = at ⇒ v = u + at. In words: the final velocity is the starting velocity plus the velocity gained in t seconds.
Worked example 1 · Keep pushing
Q. A trolley moving at 6 m/s is given a uniform acceleration of 3 m/s² for 6 s. What is its final velocity?
Known: u, a, t. Wanted: v. The distance is not involved → v = u + at.
3. s = ut + ½at²: the area under the graph
On a velocity–time graph the area under the line is the displacement. (For a steady velocity this is easy to see: 10 m/s for 5 s is a rectangle 10 high and 5 wide, and the area 50 is the 50 m covered.) For uniform acceleration the area is a rectangle plus a triangle. Watch:
Worked example 2 · From rest
Q. A trolley starts from rest and moves with a uniform acceleration of 1.5 m/s². How far does it travel in 4 s?
“From rest” ⇒ u = 0. The final velocity is not involved → s = ut + ½at², and the ut term becomes 0.
4. v² = u² + 2as: when no time is given
Combining the first two equations to remove t gives v² = u² + 2as. Use it when the question gives a distance but says nothing about time, for example a vehicle (or a bullet) that stops within a certain distance. When something stops, put v = 0 and the equation becomes 0 = u² + 2as, so a = −u² ÷ 2s (negative: a retardation).
From 0 = u² + 2as: stopping distance s = u² ÷ (2 × retardation). Double the speed and the stopping distance becomes four times as long! That is why over-speeding is so dangerous.
Worked example 3 · A motorbike stops
Q. A motorbike moving at 20 m/s is braked uniformly and stops after travelling 50 m. Find its acceleration and its retardation.
u = 20 m/s, v = 0, s = 50 m. No time given → v² = u² + 2as. Solve for a.
5. Distance–time graphs
Plot the distance moved on the vertical (y) axis against time on the horizontal (x) axis. The shape of the line tells the story:
- Horizontal line: the distance does not change, so the object is at rest.
- Straight sloping line: equal distances in equal times, so uniform velocity (zero acceleration). The steeper the line, the faster the motion: the slope = distance ÷ time = speed.
- Curve bending upwards (getting steeper): the object is speeding up (accelerating).
- Curve bending over (getting flatter): the object is slowing down.
The dots are the boat's position every second. Equal gaps give a straight line. Growing gaps give a curve that bends up, and shrinking gaps give a curve that flattens out.
6. Velocity–time graphs
Now plot velocity on the y-axis against time. Three shapes matter for this chapter:
- Horizontal line: the velocity is constant, so uniform velocity and zero acceleration.
- Straight line going up: uniform acceleration.
- Straight line going down: uniform retardation (negative acceleration), for example a ball slowed by friction.
Two rules: the slope of a v–t graph = change in velocity ÷ time = acceleration, and the area under it = displacement. To read a value, go up from the time on the x-axis to the line, then across to the y-axis.
Start: u = 8 m/s, a = −0.5 m/s². The line falls from 8 m/s to 0 at 16 s. The purple triangle shows the slope: the velocity changes by −2 m/s in every 4 s, so a = −2 ÷ 4 = −0.5 m/s². The shaded area is the distance covered so far. A body slowed by friction stops at v = 0; the line then stays on the axis.
Worked example 4 · Reading a graph
Q. The velocity–time graph of a toy car rolling on a floor is a straight line that falls from 8 m/s at t = 0 to 0 at t = 16 s. Find the acceleration of the car.
Read u and v and the time from the graph. Then a = (v − u)/t, or the slope of the line.
7. Graph detective
In the exam you may get four small graphs and be asked which one shows zero acceleration (and so, as you will learn, zero resultant force). First look at the axes: is it distance–time or velocity–time? The same straight sloping line means different things on the two kinds of graph!
A straight sloping line on a distance–time graph means uniform velocity (a = 0). A straight sloping line on a velocity–time graph means uniform acceleration (a ≠ 0). Always read the axis labels first.