2.1 - Linear Motion

Describing Motion

• Linear motion is motion along a straight line. An object is at rest when its position does not change with time.
• Distance, $d$: length of the route travelled; a scalar quantity that depends on the route.
• Displacement, $s$: shortest distance from the initial to the final position in a specified direction; a vector quantity.
• A taxi journey from Bagan Datoh to Sungkai has a distance of 83.00 km and a displacement of 57.22 km due East.
• Uniform velocity: equal changes in displacement in equal time intervals.
• Non-uniform velocity: the rate of change of displacement varies. Increasing velocity means acceleration in the direction of motion; decreasing velocity means acceleration opposite to motion.

Quantity	Relationship and S.I. unit
Speed	Rate of change of distance; $v=d/t$; $\mathrm{m}{\mathrm{s}}^{-1}$
Velocity	Rate of change of displacement; $v=s/t$; $\mathrm{m}{\mathrm{s}}^{-1}$
Acceleration	Rate of change of velocity; $a=(v-u)/t$; $\mathrm{m}{\mathrm{s}}^{-2}$

Here, $u$ is initial velocity, $v$ is final velocity and $t$ is the time interval.

Distance, Displacement and Acceleration Examples

• Radzi runs 100 m to the right, then 20 m back in 20 s:
  • Distance $=100+20=120\mathrm{m}$; displacement $=100-20=80\mathrm{m}$ to the right.
  • Speed $=120/20=6\mathrm{m}{\mathrm{s}}^{-1}$; velocity $=80/20=4\mathrm{m}{\mathrm{s}}^{-1}$ to the right.
• Muthu follows two perpendicular paths of 5 m and 12 m in 15 s:
  • Distance $=5+12=17\mathrm{m}$; displacement $=\sqrt{5^2+12^2}=13\mathrm{m}$ towards the final point.
  • Speed $=17/15=1.13\mathrm{m}{\mathrm{s}}^{-1}$; velocity $=13/15=0.87\mathrm{m}{\mathrm{s}}^{-1}$ towards the final point.
• A plane slows from 75 to 5 $\mathrm{m}{\mathrm{s}}^{-1}$ in 20 s:

Deceleration describes decreasing velocity along the direction of motion. The negative sign indicates acceleration opposite to the plane’s initial motion.

Ticker Tape and Electronic Timing

• A ticker timer connected to a 50 Hz alternating current supply makes 50 dots each second.

• One tick is the interval between adjacent dots: $1/50=0.02\mathrm{s}$. Five ticks take $0.10\mathrm{s}$; ten ticks take $0.20\mathrm{s}$.

• A ticker tape records displacement and elapsed time. For straight motion without reversing, distance and displacement have equal magnitudes.

• For a tape covering 18 cm in six ticks, $v=18/(6\times 0.02)=150\mathrm{cm}{\mathrm{s}}^{-1}$.

• If the first and seventh tick lengths are 1.2 cm and 3.6 cm:

  • $u=1.2/0.02=60\mathrm{cm}{\mathrm{s}}^{-1}$; $v=3.6/0.02=180\mathrm{cm}{\mathrm{s}}^{-1}$.
  • The interval between these velocity measurements is $(7-1)(0.02)=0.12\mathrm{s}$.
  • $a=(180-60)/0.12=1000\mathrm{cm}{\mathrm{s}}^{-2}$.

• Equal-duration tape strips form a ticker tape chart: equal lengths indicate uniform velocity; lengths increasing or decreasing by equal amounts indicate uniform acceleration or uniformly decreasing velocity.

• A photogate system and electronic timer measure motion more accurately: no attached tape introduces friction with a ticker timer, and timing resolution is 0.001 s, compared with 0.02 s for the ticker timer.

• For a trolley released from rest, electronic timing gives $a=v/t$, where $t$ is the elapsed travel time and $v$ is the final velocity measured at the photogate.

• As the trolley accelerates farther down the runway, the total travel time increases while the short interval $\Delta t$ used to measure its final velocity decreases.

Equations for Uniform Acceleration

For uniform acceleration, with $u$ and $v$ in $\mathrm{m}{\mathrm{s}}^{-1}$, $a$ in $\mathrm{m}{\mathrm{s}}^{-2}$, $t$ in seconds and $s$ in metres:

1. Rearrange $a=(v-u)/t$ to obtain $v=u+at$.
2. Multiply the average velocity, $(u+v)/2$, by time to obtain $s=(u+v)t/2$.
3. Substitute $v=u+at$ into that displacement equation to obtain $s=ut+a{t}^2/2$.
4. Square $v=u+at$: $v^2=u^2+2uat+a^2{t}^2=u^2+2a(ut+a{t}^2/2)=u^2+2as$.

Applying the Motion Equations

• School bus: starts from rest with $a=2\mathrm{m}{\mathrm{s}}^{-2}$ for 5 s; $v=0+(2)(5)=10\mathrm{m}{\mathrm{s}}^{-1}$.
• Sports car: velocity increases uniformly from 40 to 50 $\mathrm{m}{\mathrm{s}}^{-1}$ in 3 s; $s=(40+50)(3)/2=135\mathrm{m}$.
• Athlete: starts from rest and accelerates uniformly through 40 m in 8.0 s; $40=\frac{1}{2}a(8.0{)}^2$, so $a=1.25\mathrm{m}{\mathrm{s}}^{-2}$.
• Maria’s bicycle: brakes from 8 $\mathrm{m}{\mathrm{s}}^{-1}$ to rest over 2 m; $0^2=8^2+2a(2)$, so $a=-16\mathrm{m}{\mathrm{s}}^{-2}$, opposite to its initial motion.