Module 1: Mathematical Foundations · Practice · assumes Foundations 2 and 4
Five worked examples where resolving vectors is the whole method — steering into a current, holding a load still, and choosing between two good answers.
What this set is for
Resolution and relative motion are the two things from Foundation 4 that JEE returns to most often, and both appear in problems where the difficulty is in the setup rather than the arithmetic. These five build from a single required heading, through static equilibrium, to a problem with two defensible answers depending on what you are trying to achieve.
Same three-part structure: The Physics, The Mathematics, The Contemplation. An asterisk beside an example means it draws on a Foundation beyond the two assumed — noted where it occurs, because a real problem takes whatever tools it needs.
On this page
1. Flying north in a crosswind
Problem. An aircraft flies at 200 km/h relative to the air. The wind blows due east at 50 km/h relative to the ground. In what direction must the pilot point the aircraft so that it travels due north over the ground, and what is the resulting groundspeed?
The physics
The aircraft moves through the air, and the air moves over the ground. What an observer on the ground sees is the sum. If the pilot points due north, the wind will carry the aircraft east and the course will be north-east — so to go north the nose must be aimed somewhat west of north, by exactly enough that the westward part of the airspeed cancels the eastward wind. Note that the groundspeed must then be less than 200 km/h, since part of the aircraft’s effort is being spent fighting sideways rather than making progress.
Left: the heading is not the course. Right: two crossings, two different things being minimised.
The mathematics
Write east as î and north as ĵ. Let θ be the angle of the heading measured from due west toward north, so the airspeed has a westward part and a northward part:
vair/ground = 50î
vplane/air = −200 cos θ î + 200 sin θ ĵ
vplane/ground = vplane/air + vair/ground
The east–west components must cancel, since the course is due north and has no eastward part:
−200 cos θ + 50 = 0 → cos θ = 0.25
θ = cos−1(0.25) ≈ 75.5° north of west
Equivalently, about 14.5° west of north — a small correction, because the wind is only a quarter of the airspeed.
The groundspeed is whatever survives in the northward direction:
vg = 200 sin θ = 200√(1 − 0.0625) = 50√15 ≈ 193.6 km/h
Less than 200, as expected. Alternatively, since the three vectors form a right-angled triangle with the airspeed as hypotenuse: √(200² − 50²) = √37500 ≈ 193.6. Same answer, no trigonometry.
The contemplation
The aircraft points one way and travels another, and both facts are true at once. That is worth holding onto, because it is the same distinction the whole module keeps returning to: the heading is what the pilot controls, the course is what the ground sees, and neither is more real than the other. Notice also the cost of the crosswind — 6.4 km/h of speed lost to hold a straight line. Nothing was wasted on friction or drag; the loss is purely geometric, the price of spending some of a fixed speed on cancellation rather than progress.
2. A lamp on two cables
Problem. A lamp of mass 10 kg hangs from a knot held by two cables attached to the ceiling. One cable makes 30° with the ceiling, the other 60°, on opposite sides. Find the tension in each. Take g = 10 m/s².
The physics
The knot is not moving, so the three forces on it — two tensions and the weight pulling down — must add to zero. That single condition, applied to a vector equation, gives two scalar equations, and two equations determine the two unknown tensions. Before computing: the cable at 60° is steeper, so it is better placed to hold up a downward load and should carry the larger tension.
The mathematics
Put the knot at the origin, with the 30° cable going up and to the left, the 60° cable up and to the right. The weight is mg = 100 N downward.
T1 = −T1 cos 30° î + T1 sin 30° ĵ
T2 = T2 cos 60° î + T2 sin 60° ĵ
W = −100 ĵ
Horizontal balance — nothing pulls the knot sideways on balance:
−T1(√3/2) + T2(1/2) = 0 → T2 = √3 T1
Vertical balance — the upward pulls together carry the weight:
T1(1/2) + T2(√3/2) = 100
Substituting: ½T1 + (√3/2)(√3T1) = ½T1 + (3/2)T1 = 2T1 = 100
T1 = 50 N T2 = 50√3 ≈ 86.6 N
The steeper cable carries more, as predicted. A useful check: the two cables are at 30° and 60° to the horizontal on opposite sides, so they are perpendicular to each other — and indeed √(50² + 86.6²) = 100 N, the weight, exactly as two perpendicular vectors summing to it must.
The contemplation
One vector equation, ΣF = 0, became two independent scalar equations — and that is the whole reason equilibrium problems are solvable. The components do not merely help; they are what turns a statement about arrows into arithmetic with as many equations as unknowns. Notice too that the total tension, 136.6 N, exceeds the 100 N being held. Cables that pull sideways as well as up must work harder than the load requires, and as the cables approach horizontal the tensions grow without limit. That is why a washing line sags.
3. The force that holds everything still
Problem. Two forces act on a body at a point: 8 N along the positive x-axis and 6 N along the positive y-axis. Find the magnitude and direction of the third force needed to hold the body in equilibrium.
The physics
Equilibrium means the three forces sum to zero, so the third must be exactly the reverse of what the first two produce between them. Its magnitude is therefore the magnitude of their resultant, and its direction is opposite. The first two point into the first quadrant, so the third must point into the third — which is worth noting now, because the arctangent will not be able to tell.
The mathematics
F1 + F2 = 8î + 6ĵ, of magnitude √(64 + 36) = 10 N
F3 = −(8î + 6ĵ) = −8î − 6ĵ N, magnitude 10 N
The direction. The acute angle is tan−1(6/8) = 36.87°. Both components of F3 are negative, placing it in the third quadrant, so:
θ = 180° + 36.87° = 216.87° from the positive x-axis
A calculator asked for tan−1(−6/−8) returns 36.87°, pointing into the first quadrant — the exact opposite of the force required. Applying that would double the imbalance rather than cancel it.
The contemplation
Both minus signs cancelled inside the division, and with them went the only information distinguishing this force from its opposite. Foundation 3 met this with a point in the second quadrant; here it appears with a force in the third, and the consequence is more alarming — a wrong angle in a coordinate conversion misplaces a point, while a wrong angle here means applying a force in precisely the direction that makes the problem worse. The habit that prevents it costs nothing: note which quadrant the answer must lie in before computing the angle, not after.
4. Crossing a river as fast as possible
Problem. A swimmer moves at 3 m/s in still water and wants to cross a river 120 m wide, flowing at 5 m/s. What heading gets them to the far bank in the shortest time, how long does it take, and how far downstream do they land?
The physics
The crossing time depends only on how fast the swimmer closes the 120 m gap — that is, on the component of their velocity perpendicular to the bank. The current flows parallel to the bank and contributes nothing to that component, however strong it is. So to cross fastest, put the entire swimming speed into the perpendicular direction: aim straight across. Any upstream angling reduces the perpendicular part and takes longer.
Note also that the current here is faster than the swimmer. Reaching a point directly opposite is impossible — there is no heading that cancels 5 m/s of current with 3 m/s of swimming.
The mathematics
Aiming straight across gives a perpendicular velocity of the full 3 m/s:
t = 120 / 3 = 40 s
drift = 5 × 40 = 200 m downstream
The two motions are independent — the current carries the swimmer downstream at 5 m/s for the whole 40 s regardless of the crossing, and the crossing proceeds at 3 m/s regardless of the current. This is Foundation 4’s separation of components, in its simplest form.
The contemplation
The fastest crossing is achieved by ignoring the current completely, which feels wrong and is correct. It works because the current acts along a direction in which no progress is required, and a component of motion is genuinely irrelevant to a question that does not involve that direction. Students often add an upstream correction here out of instinct, and it costs them time without buying anything — the drift is not reduced by swimming partly against it if the goal is only to arrive.
5. Crossing it with the least drift *
* This one reaches outside. Minimising the drift requires differentiating, so it draws on Foundation 8 as well as 2 and 4. That is normal — a problem is set on a topic, and its solution uses whatever it needs.
Problem. The same swimmer and the same river. Now find the heading that makes the downstream drift as small as possible, and the drift achieved. Measure the heading angle α from the upstream direction along the bank, so that α = 90° means aiming straight across.
The physics
Now a different quantity is being minimised, so a different heading will win. Angling upstream reduces the net downstream speed, which helps — but it also reduces the speed across, which lengthens the crossing and gives the current more time to act. Those two effects pull against each other, and somewhere between them lies a best angle. That structure, two competing effects with an optimum in between, is the signature of a problem that needs calculus.
The mathematics
With α measured from the upstream direction, the swimmer’s velocity has an upstream part 3 cos α and an across part 3 sin α. So:
net downstream speed = 5 − 3 cos α
crossing time t = 120 / (3 sin α) = 40 / sin α
drift x(α) = 40 (5 − 3 cos α) / sin α
Check the formula before trusting it. At α = 90°, cos α = 0 and sin α = 1, giving x = 40 × 5 = 200 m — the answer from example 4. Good.
Now minimise. Differentiate the bracket by the quotient rule and set it to zero:
d/dα [(5 − 3cos α)/sin α] = [3sin²α − (5 − 3cos α)cos α] / sin²α
Setting the numerator to zero: 3sin²α + 3cos²α − 5cos α = 0
3 − 5 cos α = 0 → cos α = 3/5
The Pythagorean identity collapsed two terms into one, and what remains is startlingly simple: cos α = vswimmer / vcurrent. The numbers 120 and 40 never entered — the best heading does not depend on how wide the river is.
The drift. With cos α = 3/5 and therefore sin α = 4/5:
x = 40 (5 − 1.8) / 0.8 = 40 × 4 = 160 m
and the crossing now takes 40 / 0.8 = 50 s
So the two strategies differ: 40 s and 200 m, or 50 s and 160 m. Ten seconds bought forty metres.
The contemplation
Two correct answers to what looks like one question, and choosing between them is not physics but purpose — arriving soonest and landing nearest are different goals, and the river does not care which you want. Notice what the algebra gave back: cos α = 3/5 is the ratio of the two speeds, and the river’s width vanished on the way. When a messy expression collapses to a bare ratio, it usually means the problem had a simpler structure than its statement suggested, and it is worth stopping to ask what that structure was.
What decided each of these
In every example the resolution itself was routine. What mattered was the decision before it: which direction to resolve along, which components must cancel, which quadrant the answer lies in, and — in the last pair — which quantity was being minimised at all.
Examples 4 and 5 are the same swimmer in the same river with two different right answers. No amount of care with the algebra would tell you which to give. Only the question does.
Assumes: Foundations 2 and 4, with Foundation 8 for the final example · Type: Worked examples · Reading: 22 min · Difficulty: Intermediate to Advanced