Part 1 · Chapter 04

Motion in Two and Three Dimensions

From straight lines to curves — projectiles, circles, and shifting frames of reference

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 30 min
i What you'll learn
  • How to describe a particle's location with a position vector, and motion with vector velocity and acceleration.
  • That velocity always points tangent to the path.
  • The key idea of projectile motion: horizontal and vertical motions are independent.
  • The projectile equations, the parabolic trajectory, and the range formula.
  • Centripetal acceleration in uniform circular motion, and how velocities transform between frames of reference.
Section 4-1

What Is Physics?

Chapter 2 handled motion along one straight line. Now the motion can curve through two or three dimensions — a fighter jet banking through a tight turn, a basketball arcing toward the hoop, a plane landing on a runway. The tools are the same kinematic ideas as before, but now dressed in the vector language of Chapter 3.

Section 4-2

Position & Displacement

We locate a particle with a position vector \(r\) that reaches from the origin out to the particle:

Position vector
\[ r = x \hat{\imath} + y \hat{\jmath} + z \hat{k} \]
The coefficients x, y, z are the particle's coordinates along the three axes.

If the particle moves from \(r_{1}\) to \(r_{2}\), its displacement is the change in the position vector:

Displacement
\[ \Delta r = r_{2} - r_{1} = \Delta x \hat{\imath} + \Delta y \hat{\jmath} + \Delta z \hat{k} \]
Section 4-3

Velocity

Average velocity is displacement over time — and because displacement is a vector, so is the velocity, pointing the same way as \(\Delta r\):

Average velocity
\[ v_{\text{avg}} = \Delta r / \Delta t \]

Shrink the interval to zero and you get the instantaneous velocity, found by differentiating each component:

Instantaneous velocity
\[ v = dr/dt = v_{x} \hat{\imath} + v_{y} \hat{\jmath} + v_{z} \hat{k} \qquad \text{where} v_{x} = dx/dt, etc. \]
📐
The single most useful fact about velocity
The instantaneous velocity is always tangent to the path.

However the particle curves, its velocity arrow points along the direction of travel at that instant. (Unlike a position vector, a velocity vector doesn't reach "from here to there" — it just sits at the particle showing where it's headed.)

Section 4-4

Acceleration

When the velocity changes — in magnitude, direction, or both — the particle accelerates. Average acceleration is the change in velocity over time; instantaneous acceleration is its derivative:

Acceleration
\[ a_{\text{avg}} = \Delta v / \Delta t \qquad a = dv/dt = a_{x} \hat{\imath} + a_{y} \hat{\jmath} + a_{z} \hat{k} \]
Differentiate the velocity components to get the acceleration components.
A change in direction alone is an acceleration. Even at constant speed, a turning particle is accelerating — which is exactly what makes circular motion (Section 4-7) interesting.
Section 4-5

Projectile Motion

A projectile is launched with some initial velocity and then moves under gravity alone (ignoring air): a thrown ball, a struck golf ball, a fired cannonball — but not a powered airplane. Its acceleration is constant, \(g\) downward, with no horizontal acceleration.

⚡
The master key to all projectile problems
Horizontal and vertical motions are completely independent.

Neither affects the other. A ball dropped and a ball fired horizontally from the same height hit the ground at the same instant. This lets us split one curved 2-D problem into two easy 1-D problems: constant-velocity horizontal motion, plus free-fall vertical motion.

With launch speed \(v_{0}\) at angle \(\theta _{0}\) above the horizontal, the starting components are \(v_{0x} = v_{0} \cos \theta _{0}\) and \(v_{0y} = v_{0} \sin \theta _{0}\).

Section 4-6

Projectile Motion Analyzed

Horizontal motion has zero acceleration, so the horizontal velocity never changes:

Horizontal (constant velocity)
\[ x - x_{0} = (v_{0} \cos \theta _{0}) t \]

Vertical motion is free fall with \(a = -g\):

Vertical (free fall)
\[\begin{gathered} y - y_{0} = (v_{0} \sin \theta _{0}) t - \tfrac{1}{2} g t^{2} \\ v_{y} = v_{0} \sin \theta _{0} - g t \\ v_{y}^{2} = (v_{0} \sin \theta _{0})^{2} - 2g(y - y_{0}) \end{gathered}\]

Eliminating \(t\) between the horizontal and vertical equations gives the trajectory — and it's a parabola:

Trajectory (parabola)
\[ y = (\tan \theta _{0}) x - g x^{2} / [ 2 (v_{0} \cos \theta _{0})^{2} ] \]

The horizontal range \(R\) — the distance covered when the projectile returns to its launch height — is:

Range (launch and landing at the same height)
\[ R = (v_{0}^{2} / g) \sin 2\theta _{0} \]
Maximum when sin 2θ0 = 1, i.e. θ0 = 45°.
! Two cautions

The range formula only works when launch and landing heights are equal. When they differ — shot put, basketball — 45° is no longer the optimal angle. And these results assume no air resistance; in real air, a fly ball's range and height fall well short of the vacuum prediction.

Section 4-7

Uniform Circular Motion

A particle in uniform circular motion travels a circle at constant speed. The speed is constant, but the direction of the velocity changes continuously — so there is an acceleration. It points straight at the center and is called centripetal ("center-seeking"):

Centripetal acceleration & period
\[ a = v^{2} / r \qquad \text{(directed inward)} \qquad T = 2\pi r / v \]
r is the radius, v the speed; T is the period — the time for one full revolution.
→ Velocity

Always tangent to the circle, in the direction of motion. Constant in size, changing in direction.

⊙ Acceleration

Always points radially inward, toward the center. Constant in size, changing in direction.

Sections 4-8 & 4-9

Relative Motion

Velocity depends on who's watching. A duck flying north at 30 km/h looks stationary to a duck flying alongside. The object you attach your coordinate system to is your frame of reference — usually the ground. To translate a particle P's velocity between two frames A and B that move at constant velocity relative to each other:

Velocity transformation
\[ v_{\text{PA}} = v_{\text{PB}} + v_{\text{BA}} \]
"P relative to A" = "P relative to B" + "B relative to A." The subscripts chain together neatly. This is a vector equation — it works axis by axis in 2-D and 3-D.
🚗
A surprising invariant
All such observers measure the same acceleration: aPA = aPB.

Because the relative velocity \(v_{\text{BA}}\) is constant, its derivative vanishes — so frames moving at constant velocity relative to each other agree on accelerations, even when they disagree on velocities.

Worked Examples

Putting It to Work

1 Rescue capsule dropped from a plane

Problem. A rescue plane flies horizontally at \(v_{0} = 55.0 m/s\) at height \(h = 500 m\) and releases a capsule. How long until it hits the water, and how far ahead does it land?

Vertical first. The capsule starts with zero vertical velocity (θ0 = 0), so \(-500 = -\tfrac{1}{2}(9.8)t^{2}\) → \(t \approx 10.1 s\).

Horizontal distance
\[ x = (55.0)(10.1) \approx 555 m \]

The capsule keeps the plane's horizontal speed the whole way down, landing about 555 m ahead — directly below the plane.

2 Cannonball to a pirate ship — finding the launch angle

Problem. A cannon fires at \(v_{0} = 82 m/s\) at a ship \(R = 560 m\) away (same height). What launch angle is needed?

Solve the range equation for θ0
\[ \theta _{0} = \tfrac{1}{2} \sin ^{-1}[ gR / v_{0}^{2} ] = \tfrac{1}{2} \sin ^{-1}(0.816) \]

This gives two answers: \(\theta _{0} \approx 27^{\circ}\) and \(\theta _{0} \approx 63^{\circ}\) — a low, fast shot or a high, lofted one both land on the ship. They merge at 45° (maximum range ≈ 690 m); beyond that, the ship is safe.

3 A "top gun" turn — centripetal acceleration

Problem. A jet enters a horizontal circular turn at \((400\hat{\imath} + 500\hat{\jmath}) m/s\) and 24.0 s later exits at \((-400\hat{\imath} - 500\hat{\jmath}) m/s\). Find the acceleration in g units.

Solution. The reversed velocity means the jet completed half a circle in 24.0 s, so a full period is \(T = 48.0 s\). The constant speed is \(v = \sqrt{400^{2} + 500^{2}} \approx 640.3 m/s\). Combining \(a = v^{2}/r\) with \(T = 2\pi r/v\) gives \(a = 2\pi v/T\):

Acceleration
\[ a = 2\pi (640.3) / 48.0 \approx 83.8 m/s^{2} \approx 8.6 g \]

That's well past the ~4g where pilots grey out — a recipe for g-induced loss of consciousness.

4 A plane flying in a crosswind — relative motion

Problem. A plane has airspeed 215 km/h (its velocity relative to the wind) and must end up flying due east relative to the ground. The wind blows at 65.0 km/h, 20.0° east of north. How far must the pilot aim south of east, and what's the ground speed?

Solution. Use \(v_{\text{PG}} = v_{\text{PW}} + v_{\text{WG}}\) and demand the y-component of \(v_{\text{PG}}\) be zero:

Aim angle, then ground speed
\[\begin{gathered} \theta = \sin ^{-1}[ (65.0)(\cos 20^{\circ}) / 215 ] \approx 16.5^{\circ} south of east \\ v_{\text{PG}} = (215)(\cos 16.5^{\circ}) + (65.0)(\sin 20^{\circ}) \approx 228 \mathrm{km}/h \end{gathered}\]

The pilot points the nose 16.5° south of east to fly a straight eastward track at 228 km/h over the ground.

Review

Chapter Summary

✓ Position & displacement

\(r = x\hat{\imath} + y\hat{\jmath} + z\hat{k}\); displacement \(\Delta r = r_{2} - r_{1}\).

✓ Velocity & acceleration

\(v = dr/dt\) (tangent to path); \(a = dv/dt\). Differentiate component by component.

✓ Projectile independence

Horizontal motion (constant velocity) and vertical motion (free fall, \(a = -g\)) are independent.

✓ Trajectory & range

Path is a parabola; \(R = (v_{0}^{2}/g) \sin 2\theta _{0}\), maximum at 45° (equal heights only).

✓ Circular motion

\(a = v^{2}/r\) inward (centripetal); period \(T = 2\pi r/v\).

✓ Relative motion

\(v_{\text{PA}} = v_{\text{PB}} + v_{\text{BA}}\); all such frames agree on acceleration.

Practice

Problems

For projectiles, always split into horizontal and vertical pieces first. For relative-motion problems, write the subscript chain before plugging in numbers.

  1. A small ball rolls off a 1.20 m table and lands 1.52 m away horizontally. (a) How long is it in the air and (b) what was its speed leaving the table?
  2. A projectile is fired horizontally at 250 m/s from 45.0 m up. Find (a) the time aloft, (b) the horizontal distance to impact, and (c) the vertical speed at impact.
  3. A stone is launched at 20.0 m/s, 40.0° above horizontal. Find the horizontal and vertical displacement components at t = 1.10 s and at t = 1.80 s.
  4. A ball is thrown at a wall 22.0 m away at 25.0 m/s, 40.0° above horizontal. (a) How high up the wall does it strike, and (b) has it passed its highest point by then?
  5. A soccer ball is kicked at 19.5 m/s, 45° above horizontal. A player 55 m away starts running the instant it's kicked. What average speed lets them reach the ball just as it lands?
  6. An Earth satellite orbits 640 km up with a period of 98.0 min. Find (a) its speed and (b) its centripetal acceleration. (Add Earth's radius, ≈ 6370 km, to get the orbital radius.)
  7. A sprinter rounds a turn of radius 25 m at 10 m/s. What is the magnitude of their (centripetal) acceleration?
  8. A boat heads upstream at 14 km/h relative to the water; the river flows at 9.0 km/h relative to the ground. Find the boat's speed relative to the ground.
  9. Snow falls vertically at 8.0 m/s. At what angle from the vertical do the flakes appear to a driver moving at 50 km/h on a straight, level road?
  10. A ship must cross a 200 m river flowing east at 2.0 m/s. The boat does 8.0 m/s relative to the water, pointed 30° west of north. Find (a) the boat's velocity relative to the ground and (b) the time to cross.
Tip: at the top of a projectile's arc, the vertical velocity is zero but the horizontal velocity is unchanged — the speed there equals \(v_{0} \cos \theta _{0}\), not zero.