Equations of Motion Recap ചലനസമവാക്യങ്ങൾ
A falling coconut, a stone dropped into a well and a ball thrown straight up all move with a steady acceleration, the acceleration due to gravity. To solve the chapter's numericals (how long a fall takes, how high a stone rises, how fast it lands) you need three equations of motion. This page rebuilds them from scratch. It starts with distance, displacement, speed, velocity and acceleration, and ends with falling and upward throws.
- Units m, s, m/s, m/s² (page 2); squares and square roots (page 3).
- F = ma (page 1).
1. Distance and displacement
Distance (ദൂരം) is the total length of the path you actually travel. Displacement (സ്ഥാനാന്തരം) is the straight-line distance from where you started to where you are now, with its direction. If you walk around and come back home, your distance may be 2 km but your displacement is zero.
At the school you have walked 700 m, but the straight line from home is only 500 m (a 3-4-5 right triangle). Distance can never be less than the size of the displacement.
2. Speed and velocity
Speed = distance ÷ time, in m/s. Velocity (പ്രവേഗം) is speed in a stated direction: displacement ÷ time. “20 m/s” is a speed; “20 m/s towards Kochi” is a velocity.
- Uniform speed (സമവേഗം): equal distances in equal intervals of time, whatever the direction.
- Uniform velocity (സമപ്രവേഗം): the same speed and the same direction, so the motion is along a straight line.
A giant wheel at a temple festival turns steadily. Does a seat have uniform speed? Uniform velocity?
Uniform speed: yes, it covers equal arcs in equal times. Uniform velocity: no, its direction changes every moment. A change in velocity means there is an acceleration, even at constant speed. You will use this idea for circular motion on page 11.
3. Acceleration
When the velocity changes, the object has an acceleration (ത്വരണം): the change in velocity every second. If the velocity goes from the initial velocity u (ആദ്യപ്രവേഗം) to the final velocity v (അന്ത്യപ്രവേഗം) in time t,
“a = 2 m/s²” means the velocity grows by 2 m/s every second: 0, 2, 4, 6 m/s… If the object slows down, v is less than u and a comes out negative (also called retardation).
Worked example 1 · A bus leaves the stand
Q. A bus starts from rest at the bus stand and reaches a velocity of 20 m/s in 10 s. Find its acceleration.
“Starts from rest” means u = 0. Use a = (v − u)/t.
4. The three equations of motion
For motion along a straight line with a constant acceleration, three equations connect u, v, a, t and the displacement s:
Where do they come from? The first is just the definition of acceleration rearranged: a = (v − u)/t ⇒ v = u + at. For the second, draw the velocity–time graph: the displacement is the area under it. That area is a rectangle u × t plus a triangle ½ × t × at = ½at². The third comes from combining the first two to remove t.
Set u = 10, a = 2, t = 5: v = 20 m/s and s = 50 + 25 = 75 m, and v² = 400 = 10² + 2 × 2 × 75. Make a negative to see an object slowing down. If its velocity passes zero, the part below the axis (red) is motion backwards and counts as negative displacement.
Write down the three quantities you know and the one you want. Pick the equation that contains exactly those four. No time given or wanted? Use v² = u² + 2as.
5. Free fall: u = 0 and a = g
An object dropped (not thrown) starts with u = 0. On its way down, the Earth's pull gives it a constant acceleration g ≈ 9.8 m/s², if air resistance is small. The chapter calls this the acceleration due to gravity (ഗുരുത്വാകർഷണത്വരണം). Putting u = 0 and a = g into the three equations:
The faint coconuts are snapshots every 0.25 s. The gaps grow because the speed grows by 9.8 m/s every second. In 1 s the coconut falls 4.9 m; in 2 s it falls 19.6 m, four times as far (h ∝ t²). Air resistance is ignored.
A coconut from a 20 m palm hits the ground at almost 20 m/s (about 70 km/h) after only about 2 s. That is why you never stand under a coconut palm on a windy monsoon day!
Worked example 2 · A stone dropped into a well
Q. A stone is dropped into a deep well and hits the water 3 s later. How deep is the water surface below the point of release, and how fast is the stone moving when it hits the water? (g = 9.8 m/s²; ignore air resistance.)
Dropped ⇒ u = 0. Use h = ½gt² for the depth and v = gt for the speed.
6. Thrown straight up
Throw a ball straight up with velocity u. Take upwards as positive. Gravity pulls down, so the acceleration is a = −g: the ball loses 9.8 m/s of speed every second. Three facts make these problems easy:
- At the highest point v = 0 (for an instant the ball stops).
- Time to go up: 0 = u − g t ⇒ t = u/g. Maximum height: 0 = u² − 2gh ⇒ h = u²/2g.
- Time up = time down, and the ball comes back to your hand with the same speed u.
The graph of height against time is a symmetric arch: the rise and the fall take equal times. Switch to the Moon: g is about one sixth of the Earth's, so the same throw rises about 6 times higher and stays up about 6 times longer. You will use this on page 9.
Worked example 3 · A ball thrown up
Q. A ball is thrown vertically upwards with a velocity of 19.6 m/s (g = 9.8 m/s²). (a) How long does it take to reach the highest point? (b) What is the maximum height? (c) After how long does it come back to the thrower's hand?
At the top v = 0 and a = −g. Use v = u − gt for the time and v² = u² − 2gh for the height. Then use time up = time down.