Summary & Quick Quiz ചലനസമവാക്യങ്ങൾ
The whole chapter on one page: the eleven key boxes, every formula with its unit, the mistakes to avoid, flashcards, a scored quiz and your study checklist.
1. The chapter in one picture
Read it clockwise from the top left: displacement (with direction) divided by time is velocity; the rate of change of velocity is acceleration; both are pictured on graphs (slope and area); uniform acceleration gives the three equations of motion; and the same physics decides how far a vehicle needs to stop – road safety.
2. The key boxes of the textbook
The straight-line distance between two positions, stated with a definite direction. Symbol s, unit metre (m). Vector quantities have magnitude and direction; scalar quantities have only magnitude. – page 5
The magnitude of displacement equals the distance travelled only when the object moves along a straight path in one direction. – page 5
Displacement per unit time. A vector, in the direction of the displacement. Displacement does not depend on the path, but the whole time taken must be used. – page 6
An object moving in one direction with equal displacements in equal intervals of time has uniform velocity. If the speed or the direction changes, the velocity is non-uniform. – page 7
Change in velocity per unit time – the rate of change of velocity. A vector. Unit m/s2. – page 8
The rate of decrease of velocity = negative acceleration. Unit m/s2. a = −2 m/s2 means a retardation of 2 m/s2. – page 8
Equal changes in velocity in equal intervals of time; otherwise non-uniform acceleration (Table 2.6: 2, 2, 2 m/s2 – uniform; Table 2.7: −0.5, −1, −1, −0.25 m/s2 – non-uniform). – page 9
Choose a scale that fits the data on the graph paper. Changing the scale changes the steepness of the line, but not the value of the physical quantity it shows. – page 10
In a velocity–time graph, the displacement in a time interval equals the area under the graph for that interval. – page 10
s = ut + ½at2
v2 = u2 + 2as
Valid only for uniform acceleration. – page 11
km/h × 5/18 = m/s; m/s × 18/5 = km/h. 36 km/h = 10 m/s, 54 km/h = 15 m/s, 72 km/h = 20 m/s, 90 km/h = 25 m/s. – page 11
3. Formulas & units
📐 Formulas & units card
| Quantity | Formula | SI unit | Remember |
|---|---|---|---|
| Speed വേഗം | speed = distance/time | m/s | scalar; uses the whole path (Fig 2.4: 90/18 = 5 m/s) |
| Velocity പ്രവേഗം | v = s/t | m/s | vector; s is the displacement (Fig 2.4: 72/18 = 4 m/s, P → R) |
| Acceleration ത്വരണം | a = (v − u)/t | m/s2 | u = initial velocity, v = final velocity; negative a = retardation |
| Velocity–time relation | v = u + at | m/s | slope of the v–t graph |
| Displacement–time relation | s = ut + ½at2 | m | area of the trapezium under the v–t graph |
| Displacement–velocity relation | v2 = u2 + 2as | m2/s2 on both sides | no t in it: use it when the time is not given |
| Speed conversion | km/h × 5/18 = m/s | m/s | m/s × 18/5 = km/h; also minutes × 60 = seconds |
| Position–time graph | slope = (s₂ − s₁)/(t₂ − t₁) | m/s | slope = velocity; sloping straight line = uniform velocity |
| Velocity–time graph | slope = a; area = s | m/s2, m | straight line = uniform acceleration; horizontal line = uniform velocity |
| Train crossing a bridge | t = (train + bridge)/v | s | 200 m train, 800 m bridge, 20 m/s → 1000/20 = 50 s |
| Stopping distance | u × reaction time + u2/2a | m | double the speed → four times the braking distance (page 15) |
- Using distance instead of displacement for velocity. In Fig 2.4 the velocity is 72/18 = 4 m/s, not 90/18 = 5 m/s (that is the speed).
- Forgetting the sign or direction of displacement and velocity. Write “40 m east” or “−40 m (east +)”, not just “40 m”.
- Not converting km/h to m/s (× 5/18) or minutes to seconds (× 60) before using the equations.
- Using the equations of motion when the acceleration is not uniform. They work only for uniform acceleration.
- Writing “retardation = −2 m/s²”. Retardation is already the negative acceleration: a = −2 m/s2, retardation 2 m/s2.
- Reading the wrong scale on a graph, or thinking that a steeper line drawn with a different scale means a bigger acceleration.
- In a v–t graph, taking the height (velocity) instead of the area as the displacement.
- Forgetting that circular motion at constant speed is non-uniform velocity – the direction keeps changing.
- Train-and-bridge problems: forgetting to add the train’s own length.
4. Flashcards
Click a card to flip it.
5. Scored quiz
Answer all fourteen, then press Check my answers. Each wrong answer shows the page to revise. For number answers type only the number.
6. Your study checklist
Ticks come from “Mark this page as completed” at the bottom of each page.
Next: the model test paper ▶ – attempt it on paper with a timer first, then check with the solved paper.