Module 3: Kinematics · Worked Examples · Prerequisites: K3, K7, K8
Six problems on the parts that look easiest — averages, steady motion, falling bodies — chosen because each hides something.
Mahavakya
The simplest topics are where the wrong method is hardest to notice, because the arithmetic looks reasonable either way.
Nothing here needs calculus beyond a single integral, and no problem uses an idea past K8. They are gathered because each one has a step where a plausible route gives a wrong answer — averaging what should not be averaged, forgetting a length, or missing that a quantity is constant.
The six
1. A journey split three ways
Problem
A driver covers the first third of a journey at 40 km/h. The remaining two-thirds is covered at 80 km/h for half the time taken over it, and at 60 km/h for the other half.
Find the average speed. Then find the uniform return speed that would make the round trip average 60 km/h.
Read the split before calculating. The first division is by distance; the second is by time. K3 says those need different averages, and this problem uses both in one journey.
First third — a distance split, so work out the time it takes:
t1 = (d/3)/40 = d/120
The remaining two-thirds — a time split, so the arithmetic mean is the right one:
2d/3 = ((80 + 60)/2) trem = 70 trem
trem = d/105
ttotal = d/120 + d/105 = d/56
vavg = 56 km/h
The return leg covers the same distance, so the round trip is an equal-distance split — harmonic mean:
60 = 2(56)vR/(56 + vR) → 52vR = 3360
vR = 64.6 km/h
Notice how hard the return has to work. To lift the average from 56 to 60 — four units — the return leg must run at 64.6, which is 8.6 above the outward speed. The harmonic mean is dragged toward the slower leg, so improving it costs more than it looks.
Push it further: no return speed at all can make the round trip average 112 km/h, because 2(56)v/(56 + v) → 112 only as v → ∞. Twice the outward speed is an unreachable ceiling, and that is worth knowing before attempting such a question.
2. Two averages of the same motion
Problem
A particle crosses a stretch of length L = 100 m with a speed that grows with position: v(x) = v0(1 + x/L), where v0 = 10 m/s.
Find the average taken over time, and the average taken over position. They are not equal.
Averaged over position is the easier one — integrate the speed against x:
vspace = (1/L)∫0L v0(1 + x/L) dx = 1.5v0 = 15 m/s
Averaged over time requires the journey time first, and that needs a separation:
dt = dx/v(x) → T = (L/v0) ln 2 = 6.93 s
vtime = L/T = 10/ln 2 = 14.43 m/s
Only vtime is the average speed. Total distance over total time — that is the definition, and 14.43 m/s is the number a stopwatch and an odometer would give.
The space average is a perfectly good number that answers a different question: the mean of the speedometer reading sampled at equal intervals of road rather than of time. It is larger because the particle spends less time on the fast stretches — exactly the effect that made the harmonic mean smaller than the arithmetic one in K3, arriving here through an integral.
3. One train overtaking another
Problem
Train A is 120 m long and travels at 20 m/s. Train B is 180 m long and travels at 35 m/s on a parallel track, in the same direction. At t = 0 the front of B is 400 m behind the rear of A.
Find when the front of B draws level with the rear of A, when B has completely cleared A, and how far B travels over the ground in that time.
Work in B’s frame. A recedes at 15 m/s, and the two questions differ only in how much relative distance must be covered.
first contact: t1 = 400/15 = 26.7 s
To clear completely, B must also pass both lengths:
t2 = (400 + 120 + 180)/15 = 700/15 = 46.7 s
DB = 35 × 46.7 = 1633 m over the ground
Both lengths, not one. B’s front must travel from the rear of A to beyond A’s front — that is A’s length — and then B’s own tail must follow. Forgetting either is the standard error, and either omission still produces a plausible answer.
The last part deliberately breaks frames. The times come from the relative frame; the distance over the ground does not. DB = 15 × 46.7 = 700 m is the relative displacement, and it is not what was asked.
4. Two cars that never meet
Problem
Car 1 drives east at 12 m/s and is 100 m west of a crossroads. Car 2 drives north at 16 m/s and is 120 m south of it. Both start at t = 0.
Find their closest approach, and when it happens.
S²(t) = (−100 + 12t)² + (−120 + 16t)² = 400t² − 6240t + 24400
Minimise the square — the square root is monotonic and only complicates it:
800t − 6240 = 0 → t = 7.8 s
Smin = √64 = 8.0 m
They miss by eight metres. A collision would need Smin = 0, and the discriminant of 400t² − 6240t + 24400 tells you at once that it never reaches zero.
Minimise the square, never the distance. Differentiating √(…) means carrying a chain rule through a square root for no gain — and since squaring preserves order for non-negative quantities, the minimum sits at the same instant.
5. Two balls, one collision
Problem
A ball is dropped from a 100 m tower at t = 0. Two seconds later a second ball is thrown up from the base at 30 m/s. Take g = 9.8 m/s².
Find when and where they collide, and their relative speed at that instant.
y1 = 100 − 4.9t² y2 = 30(t−2) − 4.9(t−2)²
Setting them equal, the 4.9t² terms cancel and the equation is linear:
100 = 49.6t − 79.6 → t = 3.62 s
y = 100 − 4.9(3.62)² = 35.8 m
The relative speed at impact: v1 = −35.5 m/s and v2 = +14.1 m/s, so they close at 49.6 m/s.
That 49.6 never changes, and the last part need not have been computed at all. Both balls share the same acceleration, so their relative acceleration is zero and the relative velocity is constant from the moment the second is launched. Its value is 30 + 9.8(2) = 49.6 m/s — the launch speed plus whatever the first ball had picked up in its two-second head start.
This is K13’s central result, and it also explains why the quadratic collapsed: a linear equation in t is what zero relative acceleration looks like. Seeing that in advance turns the whole problem into one line.
6. A stone past a window
Problem
A stone dropped from rest takes 0.15 s to pass a window 2.45 m tall. Take g = 9.8 m/s².
From what height above the top of the window was it released?
Treat the window as an interval, not a position. The stone enters its top at some unknown speed and covers 2.45 m in 0.15 s:
2.45 = vtop(0.15) + ½(9.8)(0.15)²
2.45 − 0.110 = 0.15vtop → vtop = 15.6 m/s
H = vtop²/(2g) = 243.4/19.6 = 12.4 m
The 0.110 m matters more than it looks. Dropping it — treating the stone as travelling at a steady vtop across the window — gives 16.3 m/s and a height of 13.6 m, about 10% high. The window is short enough that the approximation is tempting and wrong.
The two stages are also worth naming: the window gives a velocity, and only then does v² = 2gH give a height. Trying to reach H in one step means writing an equation with two unknowns.
What the six have in common
Every one of these could be attempted with a formula and produce a plausible wrong number. Average the two speeds in the first and you get 60 instead of 56. Forget a train length in the third and you get 40 s instead of 46.7. Drop the ½gt² term in the last and you are 10% out.
None of those errors announces itself. There is no dimensional inconsistency, no negative time, no impossible geometry — just a number of the right size and the wrong value. That is what makes these topics deceptive: the checks that catch mistakes elsewhere do not fire here.
What does catch them is asking, before any arithmetic, what exactly is being divided by what, and what has to move for this event to happen. Both are questions about the situation rather than about the formula.
Carried forward
- Equal distances take the harmonic mean; equal times take the arithmetic. Check which split a problem uses — it may use both.
- No return speed can raise a round-trip average above twice the outward speed.
- The average speed is the time average. A position average is a different, legitimate quantity answering a different question.
- Overtaking needs both lengths cleared, not one.
- Times may come from a relative frame; ground distances do not.
- Minimise the square of a separation, never the separation.
- Equal accelerations mean constant relative velocity — and a collision equation that turns out linear.
Prerequisites: K3, K7, K8 · Working time: 40 min · Level: JEE Main, with two Advanced parts