Motion in a Straight Line

A passenger sitting in a moving train is at rest relative to the train and moving relative to the platform. Every statement about position, velocity or rest needs a frame of reference: an origin, a positive direction and a clock. In one dimension, a single signed coordinate \(x\) is enough.

We treat the body as a point object when its size is small compared with the distance it moves.

• Distance is a scalar and never negative. Displacement is a vector and can be positive, negative or zero.
• \(|\text{displacement}|\le\text{distance}\). They are equal only when the body never reverses.
• Distance never decreases with time. Displacement can decrease.
• Zero displacement with non-zero distance is common (a round trip). Zero distance forces zero displacement.

For a position function \(x(t)\): find where \(v=dx/dt=0\) inside the interval, split there, and add the size of each piece.

Although this chapter is one-dimensional, questions often compare distance and displacement on a circle of radius \(R\). Half a circle: distance \(\pi R\), displacement \(2R\). Quarter circle: \(\pi R/2\) and \(\sqrt2R\). Full circle: \(2\pi R\) and zero.

Equal time intervals at speeds \(v_1\) and \(v_2\): total distance \(=(v_1+v_2)t\) in time \(2t\), so \(\bar v=\dfrac{v_1+v_2}{2}\) (arithmetic mean).

Equal distances \(d\) each: times are \(d/v_1\) and \(d/v_2\), so \(\bar v=\dfrac{2d}{d/v_1+d/v_2}=\dfrac{2v_1v_2}{v_1+v_2}\) (harmonic mean). The slower leg takes more time and pulls the average down.

For three equal distances: \(\bar v=\dfrac{3v_1v_2v_3}{v_1v_2+v_2v_3+v_3v_1}\). The harmonic mean is always less than or equal to the arithmetic mean.

Average velocity between two times is the slope of the chord joining the two points. Instantaneous velocity is the slope of the tangent. As the interval shrinks, the chord turns into the tangent: \(v=\lim_{\Delta t\to0}\Delta x/\Delta t=dx/dt\).

For constant acceleration only, the average velocity equals \((u+v)/2\), and it equals the instantaneous velocity at the middle of the time interval.

Average speed \(\ge|\text{average velocity}|\), because distance \(\ge|\text{displacement}|\). For a round trip, average velocity is zero but average speed is not. Instantaneous speed always equals the magnitude of instantaneous velocity, because over a tiny interval there is no time to turn back.

Rule: speed increases when \(v\) and \(a\) have the same sign and decreases when they have opposite signs.

Average acceleration \(=\Delta v/\Delta t\), the slope of the chord on a v–t graph. Instantaneous acceleration \(a=dv/dt=d^2x/dt^2\), the slope of the tangent. On an x–t graph, acceleration shows as curvature: bending up means \(a\gt0\), bending down means \(a\lt0\), straight means \(a=0\).

Velocity is a vector, so a change of direction alone is an acceleration. A body that reverses from +10 m/s to −10 m/s in 4 s has average acceleration −5 m/s² even though its speed is the same at both ends.

"Retardation" or "deceleration" means acceleration opposite to velocity, so the speed falls. It does not mean a negative sign. With a different choice of positive direction, the same retardation can have a positive sign.

1. The v–t graph is a straight line of slope \(a\) and intercept \(u\): \(v=u+at\).
2. Displacement is the area under it: a rectangle \(ut\) plus a triangle \(\tfrac12(at)t\). So \(s=ut+\tfrac12at^2\).
3. The area is also a trapezium: \(s=\tfrac12(u+v)t\). Put \(t=(v-u)/a\) to get \(v^2=u^2+2as\).

Calculus gives the same results: integrate \(a=dv/dt\) once for \(v\), and \(v=dx/dt\) again for \(s\).

A driver first reacts for a time \(t_r\), during which the car keeps its speed. Then the brakes give retardation \(a\).

Without reaction time, stopping distance \(\propto u^2\) and stopping time \(\propto u\). Doubling speed makes the braking distance four times as long.

Starting from rest with constant \(a\):

• Distances in 1, 2, 3 … seconds (total): 1 : 4 : 9 …
• Distances in the 1st, 2nd, 3rd … seconds: 1 : 3 : 5 … (odd numbers)
• Times to cover successive equal distances: \(1:(\sqrt2-1):(\sqrt3-\sqrt2)\) …
• Velocities after successive equal distances: \(1:\sqrt2:\sqrt3\) …

1. Displacement in \(n\) s: \(un+\tfrac12an^2\).
2. Displacement in \((n-1)\) s: \(u(n-1)+\tfrac12a(n-1)^2\).
3. Subtract: \(u+\tfrac12a\left[n^2-(n-1)^2\right]=u+\tfrac a2(2n-1)\).

Written fully, \(s_n=u(1\,\text{s})+\tfrac a2(2n-1)(1\,\text{s})^2\). The "1 s" is the length of the interval, so the formula is dimensionally fine.

• From rest: \(s_1:s_2:s_3\ldots=1:3:5\ldots\) (Galileo's odd numbers).
• Consecutive seconds always differ by \(a\times(1\,\text{s})^2\): \(s_{n+1}-s_n=a\). So two such distances give the acceleration directly.
• From rest, the ratio of the distance in the nth second to the distance in n seconds is \((2n-1)/n^2\).

With up positive, \(s_n=u-\tfrac g2(2n-1)\). In the last second before reaching the top, any upward throw rises \(g/2=5\) m. That second mirrors the first second of a free fall from rest. If the top falls inside the nth second, split the second at the top to get the distance.

• An x–t curve that doubles back in time (two positions at one instant).
• A vertical segment on x–t (infinite velocity) or on v–t (infinite acceleration).
• A distance–time graph that decreases, or a speed–time graph below the axis.
• Total path length less than the size of the displacement.

From v–t to x–t: where v is positive, x rises; where v is zero, x has a flat point; a straight v–t line gives a parabolic x–t piece. From v–t to a–t: read the slope of each straight piece. From a–t to v–t: add the area of each piece to the starting velocity.

A ball dropped onto a hard floor that bounces back to the same height gives a sawtooth v–t graph: straight lines of slope −g, with sudden jumps from −v to +v at each bounce.

1. a = f(t): \(v=u+\int_0^tf(t)\,dt\), then \(x=x_0+\int_0^tv\,dt\).
2. a = f(x): \(v\,dv=f(x)\,dx\Rightarrow\tfrac12v^2-\tfrac12u^2=\int_{x_0}^{x}f(x)\,dx\).
3. a = f(v): for time, \(\int\dfrac{dv}{f(v)}=t\); for distance, \(\int\dfrac{v\,dv}{f(v)}=x\).

Maximum or minimum velocity occurs where \(a=0\).

By the chain rule, \(a=\dfrac{dv}{dt}=\dfrac{dv}{dx}\dfrac{dx}{dt}=v\dfrac{dv}{dx}\). Also \(v\dfrac{dv}{dx}=\dfrac{d}{dx}\left(\tfrac12v^2\right)\). So on a graph of \(v^2\) against \(x\), the slope is \(2a\). On a v–x graph, \(a\) at a point is the velocity times the slope there.

• \(v=k\sqrt x\) means \(v^2=k^2x\): constant acceleration \(a=k^2/2\).
• \(v=kx\) gives \(a=k^2x\): acceleration grows with distance.
• \(a=-kv\) (drag proportional to speed): \(v=ue^{-kt}\). The body never quite stops, but its total distance is finite, \(u/k\).
• \(a=-kv^2\): \(1/v=1/u+kt\).
• Position given as \(t=\alpha\sqrt x+\beta\): rewrite as \(x=\left(\frac{t-\beta}{\alpha}\right)^2\), then differentiate.

Take up as positive, so \(a=-g\). For launch speed \(u\):

• Time to top \(=u/g\); maximum height \(H=u^2/2g\); total time back \(=2u/g\).
• The ball passes every height with the same speed going up and coming down: \(v=\sqrt{u^2-2gh}\).
• It returns to the launch point with speed \(u\), pointing down.
• If it is at height \(h\) at times \(t_1\) and \(t_2\): \(t_1+t_2=2u/g\) and \(t_1t_2=2h/g\). So \(u=g(t_1+t_2)/2\) and \(h=gt_1t_2/2\).

An object released from a rising balloon keeps the balloon's upward velocity at that instant. It first rises, then falls. Write one equation with the launch point as origin: \(-H=ut-\tfrac12gt^2\), where \(H\) is the height of release. Solve the quadratic and keep the positive root.

\(h=\tfrac12gt^2\), \(v=gt=\sqrt{2gh}\). Distances in successive seconds: 5, 15, 25, 35 m (ratio 1 : 3 : 5 : 7) with \(g=10\) m/s². Time to fall varies as \(\sqrt h\) and impact speed varies as \(\sqrt h\). Without air resistance, mass does not matter.

\(v_{AB}=v_A-v_B\) is the velocity of A as seen by B. Also \(v_{BA}=-v_{AB}\). With a common positive direction:

• Same direction: relative speed \(=|v_A-v_B|\).
• Opposite directions: relative speed \(=|v_A|+|v_B|\).

Crossing trains: to pass completely, one train must move its whole length plus the other's length relative to the other. Time \(=\dfrac{L_1+L_2}{v_{\text{rel}}}\).

Use relative initial separation, relative initial velocity and relative acceleration in one equation: \(s_{\text{rel}}=u_{\text{rel}}t+\tfrac12a_{\text{rel}}t^2\).

• Two bodies in free fall have \(a_{\text{rel}}=0\), so their relative velocity stays constant. A stone dropped from a tower and another thrown up from the ground meet after \(t=H/u\).
• Chase: a car at constant \(v\) passes a jeep at rest that starts with acceleration \(a\). They meet when \(vt=\tfrac12at^2\), so \(t=2v/a\). The jeep then has speed \(2v\).
• Avoiding a collision: a faster car behind a slower one brakes. It avoids hitting it if \(u_{\text{rel}}^2\le2a\,d\), where \(d\) is the gap.

For an object dropped inside a lift, use the lift as the frame. If the lift moves at constant velocity, the object falls as if the lift were at rest. If the lift accelerates upward at \(a\), the object falls relative to the floor with \(g+a\). If downward at \(a\), with \(g-a\).