Motion in a Plane

• Average velocity \(= \Delta\vec r/\Delta t\). It points along the chord joining the two positions.
• Instantaneous velocity \(\vec v = d\vec r/dt = \dfrac{dx}{dt}\hat i + \dfrac{dy}{dt}\hat j\). As Δt shrinks, the chord turns into the tangent, so \(\vec v\) is always tangent to the path.
• Acceleration \(\vec a = d\vec v/dt\). It need not be along the path or along \(\vec v\).
• Average speed (path length ÷ time) is never less than the magnitude of average velocity. They are equal only for straight-line motion without turning back.

Split \(\vec a\) into a part along \(\vec v\) and a part across it.

• \(\vec v\cdot\vec a > 0\): speed increasing. \(\vec v\cdot\vec a < 0\): speed decreasing. \(\vec v\cdot\vec a = 0\) at every instant: speed constant (only the direction changes).
• The along-velocity part has size \(a_t = \vec v\cdot\vec a / v\). The across part bends the path.
• With constant \(\vec a\): if \(\vec u\) is zero or parallel to \(\vec a\), the path is a straight line. Otherwise it is a parabola.

1. Write x(t) and y(t) separately.
2. Differentiate once for \(v_x, v_y\); again for \(a_x, a_y\).
3. Substitute the time only at the end, then combine with Pythagoras.
4. For the path, eliminate t between x and y. For example \(x = 3t,\ y = 4t^2\) gives \(y = 4x^2/9\), a parabola.

Take +y upward. Initial components: \(u_x = u\cos\theta\), \(u_y = u\sin\theta\). Accelerations: \(a_x = 0\), \(a_y = -g\).

1. Time to the top. \(v_y = u\sin\theta - gt = 0\) gives \(t_{\rm up} = u\sin\theta/g\).
2. Time of flight. On level ground the fall mirrors the rise, so \(T = 2t_{\rm up} = 2u\sin\theta/g\). (Or set y = 0 in \(y = u\sin\theta\,t - \tfrac12gt^2\).)
3. Maximum height. \(v_y^2 = u_y^2 - 2gy\) with \(v_y = 0\): \(H = u^2\sin^2\theta/2g\).
4. Range. Horizontal speed is constant, so \(R = u\cos\theta \cdot T = 2u^2\sin\theta\cos\theta/g = u^2\sin2\theta/g\).

• \(v_x = u\cos\theta\) always; \(v_y = u\sin\theta - gt\). The direction is \(\tan\varphi = v_y/v_x\).
• At the top, \(v = u\cos\theta\) (not zero), and kinetic energy is \(K\cos^2\theta\), where K is the launch KE.
• At height h on the way up or down, \(v^2 = u^2 - 2gh\), whatever the angle.
• Over the full flight on level ground, \(\Delta\vec v = 2u\sin\theta\) downward and \(\Delta\vec p = 2mu\sin\theta\) downward. Over half the flight it is \(u\sin\theta\) downward.
• Velocity and acceleration are perpendicular only at the top.

• \(R_{\max} = u^2/g\) at θ = 45°. Then \(H = R_{\max}/4\).
• \(\dfrac{H}{R} = \dfrac{\tan\theta}{4}\), so \(R = 4H\cot\theta\). R = H when tan θ = 4 (θ ≈ 76°).
• \(H = gT^2/8\). This links height to flight time with no angle or speed needed.
• Scaling at a fixed angle: T ∝ u, while H and R ∝ u². Doubling the speed doubles T but makes H and R four times larger.

1. From the horizontal motion, \(t = x/(u\cos\theta)\).
2. Substitute into \(y = u\sin\theta\,t - \tfrac12gt^2\): \(y = x\tan\theta - \dfrac{g x^2}{2u^2\cos^2\theta}\).
3. Factor out \(x\tan\theta\): \(y = x\tan\theta\left(1 - \dfrac{gx}{2u^2\sin\theta\cos\theta}\right) = x\tan\theta\left(1 - \dfrac{x}{R}\right)\).

Reading a given path \(y = bx - cx^2\): \(\tan\theta = b\), range \(R = b/c\), and maximum height \(H = b^2/4c\) (at x = R/2).

For launch angles θ and 90° − θ at the same speed u:

• Ranges are equal.
• Heights: \(H_1 = u^2\sin^2\theta/2g\), \(H_2 = u^2\cos^2\theta/2g\). So \(H_1 + H_2 = u^2/2g\) and \(H_1H_2 = R^2/16\), giving \(R = 4\sqrt{H_1H_2}\).
• Times: \(T_1T_2 = \dfrac{4u^2\sin\theta\cos\theta}{g^2} = \dfrac{2R}{g}\).
• At 45° the two angles coincide and the range is the maximum, \(u^2/g\).

To check whether a ball clears a wall at horizontal distance d, put x = d into the trajectory and compare y with the wall’s height. This avoids finding the time separately. The form \(y = x\tan\theta(1 - x/R)\) is the quickest when R is already known.

Launch horizontally at speed u from height h, with the origin at the launch point and y measured downward.

1. x = ut and y = ½gt². Eliminating t: \(y = \dfrac{g x^2}{2u^2}\), half of a parabola.
2. At impact, \(v_x = u\) and \(v_y = \sqrt{2gh}\), so \(v = \sqrt{u^2 + 2gh}\).
3. Angle of velocity below the horizontal: \(\tan\varphi = gt/u\). Angle of the displacement: \(\tan\alpha = y/x = gt/2u\). Hence \(\tan\varphi = 2\tan\alpha\) at every instant.

Keep one origin at the launch point and one sign convention. For a launch at θ above the horizontal from height h:

Solve the quadratic and keep the positive root. The horizontal distance is then \(u\cos\theta\,t\). The impact speed is \(\sqrt{u^2 + 2gh}\) for any launch angle, because energy conservation does not care about direction.

A packet released from a plane flying horizontally at constant velocity keeps the plane’s horizontal velocity. With no drag it stays vertically below the plane while it falls. A passenger sees it fall straight down; a person on the ground sees a parabola. It lands ahead of the point above which it was released, by a distance \(u\sqrt{2h/g}\).

1. x′ along the incline (up the slope positive), y′ perpendicular to it.
2. Initial velocity: \(u\cos\theta\) along x′, \(u\sin\theta\) along y′.
3. Acceleration: \(-g\sin\alpha\) along x′ (when thrown up the slope), \(-g\cos\alpha\) along y′.
4. Landing: y′ = 0, so \(u\sin\theta\,T - \tfrac12g\cos\alpha\,T^2 = 0\), giving \(T = 2u\sin\theta/(g\cos\alpha)\). The same T holds for a throw down the slope.
5. Range: \(R = u\cos\theta\,T - \tfrac12g\sin\alpha\,T^2\), which simplifies to \(2u^2\sin\theta\cos(\theta+\alpha)/(g\cos^2\alpha)\). Down the slope, \(+\tfrac12g\sin\alpha T^2\) gives \(\cos(\theta-\alpha)\).

Setting α = 0 recovers the level-ground results. That is a good check on any incline formula.

• Up the incline: \(R_{\max} = \dfrac{u^2}{g(1+\sin\alpha)}\) at \(\theta = 45^\circ - \alpha/2\) from the incline.
• Down the incline: \(R_{\max} = \dfrac{u^2}{g(1-\sin\alpha)}\) at \(\theta = 45^\circ + \alpha/2\).
• The best direction bisects the angle between the incline and the vertical.
• Strikes the incline perpendicularly (thrown up the slope): the velocity along the incline must be zero at landing, \(u\cos\theta = g\sin\alpha\,T\). This gives \(\cot\theta = 2\tan\alpha\).

A ball is thrown horizontally at u from the top of an incline of angle α. It lands when the drop equals the horizontal distance times tan α:

1. \(\tfrac12gt^2 = ut\tan\alpha\), so \(t = 2u\tan\alpha/g\).
2. Horizontal distance \(x = ut = 2u^2\tan\alpha/g\).
3. Distance along the incline \(L = x/\cos\alpha = 2u^2\tan\alpha/(g\cos\alpha)\).

Here θ = α measured from the incline, and the general formula gives the same answer.

The rain’s velocity as seen by a moving person is \(\vec v_{R/M} = \vec v_R - \vec v_M\). Hold the umbrella against this relative velocity.

• Rain falling vertically at \(v_R\), person walking at \(v_M\): the rain appears to come from the front at \(\tan\theta = v_M/v_R\) to the vertical. Tilt the umbrella forward.
• If the rain itself is slanted, the person can make it appear vertical by walking with the rain’s horizontal component.
• Two-condition questions (“appears vertical at 3 km/h, at 45° at 6 km/h”) are solved by writing the rain’s velocity as \(a\hat i - b\hat j\) and using each condition once.

Let the river (width d) flow at w along +x and the swimmer’s speed relative to water be u, aimed at angle θ upstream from the straight-across direction.

1. Across component: \(u\cos\theta\), so the crossing time is \(t = d/(u\cos\theta)\).
2. Along the bank: \(w - u\sin\theta\), so the drift is \(x = (w - u\sin\theta)\,d/(u\cos\theta)\).
3. θ = 0: least time \(d/u\), drift \(wd/u\).
4. \(\sin\theta = w/u\): zero drift (needs u > w), time \(d/\sqrt{u^2 - w^2}\).
5. If u < w the swimmer cannot reach the point opposite. The least drift is \(d\sqrt{w^2-u^2}/u\), found with \(\sin\theta = u/w\).

For two bodies, work in the frame of one of them. The relative velocity tells you the direction of approach; closest approach and “will they collide?” questions become straight-line problems. Two projectiles launched at the same time have the same acceleration g, so their relative acceleration is zero. Each sees the other move in a straight line at constant velocity.

1. In a short time Δt the particle turns through Δθ. Its speed stays v, so \(\vec v_1\) and \(\vec v_2\) have equal length and differ in direction by Δθ.
2. Draw them from one point: \(|\Delta\vec v| = 2v\sin(\Delta\theta/2) \approx v\,\Delta\theta\) for small Δθ.
3. Divide by Δt: \(a = v\,\dfrac{\Delta\theta}{\Delta t} = v\omega = \dfrac{v^2}{r}\), using ω = v/r.
4. As Δθ → 0, \(\Delta\vec v\) becomes perpendicular to \(\vec v\) and points to the centre. That is why it is called centripetal (centre-seeking).

• Constant: speed, kinetic energy, angular speed ω, the magnitude of acceleration, the magnitude of momentum.
• Changing: velocity, momentum and acceleration, all as vectors, because their directions turn.
• Over half a revolution: displacement 2r, distance πr, \(|\Delta\vec v| = 2v\). Average velocity \(= 2r/(T/2) = 4r/T\); average speed \(= 2\pi r/T\).
• Over a full revolution: displacement, \(\Delta\vec v\) and average velocity are all zero.

ω = 2π/T = 2πf. So \(a_c = \omega^2r = 4\pi^2r/T^2 = 4\pi^2f^2r\). For the hands of a clock: second hand T = 60 s, minute hand T = 3600 s, hour hand T = 43 200 s. The ratio of the angular speeds of the minute and hour hands is 12 : 1.

The total acceleration makes angle β with the inward radius, where \(\tan\beta = a_t/a_r\).

• Speed increasing: the acceleration leans forward of the radius (toward the velocity).
• Speed decreasing: it leans backward.
• Uniform motion: β = 0, purely radial. A straight line (r → ∞): purely tangential.

The angle between \(\vec a\) and \(\vec v\) is 90° − β when speeding up and 90° + β when slowing down.

With constant \(a_t\) from rest: \(v = a_t t\), so \(a_r = a_t^2t^2/r\) grows with time. In terms of the angle covered, \(v^2 = 2a_t(r\theta)\) gives

So the two parts are equal after half a radian, and after one full revolution (θ = 2π) the radial part is \(4\pi a_t\). The total acceleration changes direction continuously even though \(a_t\) is constant.

For constant angular acceleration α the straight-line equations carry over with θ for s, ω for v and α for a:

• \(\omega = \omega_0 + \alpha t\)
• \(\theta = \omega_0t + \tfrac12\alpha t^2\)
• \(\omega^2 = \omega_0^2 + 2\alpha\theta\)

Number of revolutions = θ/2π. The links to linear motion are \(v = r\omega\) and \(a_t = r\alpha\).