Module 3: Kinematics · Worked Examples · Advanced · Prerequisites: K9, K11, K12, K15
Six problems where the answer is a boundary — how far, how high, how fast at most, and what cannot be reached at all.
Mahavakya
A limit is not the end of a calculation. It is a fact about the motion, and often the only fact worth having.
These six span variable acceleration, motion in two dimensions and projectiles — the three parts with no worked-example set of their own. They are gathered here because they share a shape: in each, the interesting quantity is an extreme rather than a value at some particular instant.
The six
1. A pull that weakens with distance
Problem
A particle released from rest at x0 is drawn toward the origin with a = −k/x².
Find its speed as a function of position, say what happens as it approaches the origin, and find the time to fall half way.
No time appears in the first two questions, so use v dv = a dx:
½v² = ∫ −k/x² dx = k/x − k/x0
v = −√(2k(1/x − 1/x0))
Negative because the particle moves toward decreasing x. As x → 0 the term 1/x diverges, so the speed grows without limit.
For the time, separate and integrate — the substitution x = x0 cos²θ turns it into a standard form:
T = x03/2(π + 2) / (4√(2k))
An infinite speed is a signal, not a result. No real body reaches one — what the divergence tells you is that the model has stopped applying. A point mass has no size, so nothing stops the particle arriving at the centre; a real attractor has a surface, and the motion ends there.
This is the same reading as K15’s ladder: when a constraint drives a quantity to infinity, look for the physical feature the model left out.
2. A pull that strengthens with distance
Problem
A particle leaves the origin at v0 = 12 m/s against a retardation a = −kx² with k = 0.5 in SI units.
How far does it get?
½(v² − v0²) = −k ∫0x x′² dx′ = −kx³/3
Stopping means v = 0:
xmax = (3v0²/2k)1/3 = ∛432 ≈ 7.56 m
Compare the two problems. Both are a(x), both use the same identity, and both end at a limit. But example 1’s limit is a speed that diverges, and this one’s is a distance that is finite — because a retardation growing as x² wins the race against the kinetic energy available.
The cube root is the tell. Doubling the launch speed multiplies the reach by only 22/3 ≈ 1.59, not by four as it would under constant retardation.
3. Splitting the acceleration in two
Problem
A particle moves with r(t) = 2t² i + (t³ − 4t) j metres.
At t = 2 s, split the acceleration into its tangential and normal parts.
v = 4t i + (3t² − 4) j → v(2) = 8i + 8j, |v| = 8√2 ≈ 11.31 m/s
a = 4 i + 6t j → a(2) = 4i + 12j, |a| = √160
The tangential part is the projection of a onto the direction of v:
at = (a · v)/|v| = (32 + 96)/11.31 = 11.31 m/s²
an = √(|a|² − at²) = √(160 − 128) = 5.66 m/s²
check: 128 + 32 = 160 = |a|² ✓
Never compute an directly when you can get it by subtraction. Finding the normal component from geometry means locating the centre of curvature; taking it as √(|a|² − at²) needs only Pythagoras, and the check falls out for free.
And an = 5.66 gives the radius of curvature immediately: R = v²/an = 128/5.66 ≈ 22.6 m.
4. Firing up a slope
Problem
A projectile is launched at u = 30 m/s up a plane inclined at β = 30°, at an angle α to the plane. Take g = 10 m/s².
Which α maximises the range along the plane, and what is that range?
Rotate the axes. With x′ along the slope and y′ perpendicular to it, gravity acquires two components:
ax′ = −g sin β ay′ = −g cos β
The flight ends when y′ returns to zero, giving T = 2u sin α/(g cos β)
Substituting into x′ and simplifying with 2 sin A cos B = sin(A+B) + sin(A−B):
R = u²[sin(2α + β) − sin β] / (g cos² β)
Only sin(2α + β) depends on α, and it is largest at 1:
2α + β = 90° → α = (90° − β)/2 = 30°
Rmax = u²(1 − sin β)/(g cos² β) = u²/(g(1 + sin β)) = 900/15 = 60 m
The optimum angle bisects the slope and the vertical. On level ground β = 0 and α = 45°, the familiar result. Tilt the ground and the best angle tilts by half as much.
Notice how much the identity did. Written as sin(2α + β), the optimisation is immediate; left as sin α cos(α + β) it needs calculus.
5. The same slope, in symbols
Problem
A projectile is fired at v0 = 40 m/s at α = 60° to the horizontal, up a ramp inclined at β = 30°. Take g = 9.8 m/s².
Find the time of flight and the range along the ramp.
The angle to the ramp is α − β = 30°. Everything from example 4 applies with that substitution:
T = 2v0 sin(α − β)/(g cos β) = 2(40)(0.5)/(9.8 × 0.866) = 4.71 s
R = 2v0² sin(α − β) cos α / (g cos² β) = 108.8 m
Check it the long way once. Working in rotated components with ax′ = −g sin β and T = 4.71 s gives x′ = 108.8 m — the same answer by a different route, which is worth doing whenever a result arrives through a trigonometric identity.
The trap is the angle. Here α is measured from the horizontal, not from the ramp. Feeding 60° into example 4’s formula instead of 30° gives a plausible number and a wrong one.
6. Everything the projectile can reach
Problem
A launcher fires at a fixed u = 20 m/s but any angle. Take g = 10 m/s².
Find the boundary of the region it can reach, then decide whether a target at (30, 5) is inside it — and if so, at what angles.
Write the trajectory as a quadratic in T = tan θ:
(gx²/2u²)T² − xT + (y + gx²/2u²) = 0
A point is reachable exactly when this has a real root. Setting the discriminant to zero gives the boundary:
yenv = u²/(2g) − gx²/(2u²)
The target lies below the boundary, so two angles reach it.
at x = 30: yenv = 20 − 10(900)/800 = 8.75 m
The target sits at 5 m, below 8.75 — reachable.
Solving 9T² − 24T + 13 = 0: θ = 62.4° and 37.1°
The discriminant answers a question the trajectory equation cannot. Asking “can I hit this?” is not asking “where does it land?” — and the number of solutions is the answer: two angles below the boundary, one on it, none above.
The two angles are not complementary here. That rule holds only for targets on the ground, where y = 0; lift the target and the pair shifts. 62.4° and 37.1° sum to 99.5°, not 90°.
What the six have in common
In every one, the answer was a limit: a speed that diverges, a distance that does not, an acceleration split into the largest part along the motion and the rest across it, an angle that maximises a range, a boundary beyond which nothing can be reached.
Limits behave differently from ordinary answers. An infinite one usually means the model has run out rather than the physics has — example 1’s diverging speed, like K15’s ladder leaving the wall. A finite one is often more informative than any particular value: knowing the launcher cannot reach 9 m at 30 m out settles more questions than knowing where one shot landed.
And in the last example the limit arrives from a discriminant — the same object that told K7 a collision never happens. Counting solutions is a physical question wearing algebraic clothes.
Carried forward
- When no time appears in the question, reach for v dv = a dx.
- A diverging answer is a statement about the model, not about the motion.
- Get an by subtraction from |a|, never by locating the centre of curvature.
- On a slope, rotate the axes; the optimum launch bisects the slope and the vertical.
- Check which line an angle is measured from before substituting it.
- A discriminant answers can this be done? — a different question from what happens?
Prerequisites: K9, K11, K12, K15 · Working time: 50 min · Level: JEE Advanced