By the end of the subtopic, learners should be able to:
define displacement, speed, velocity and acceleration
distinguish between, distant and displacement, speed and velocity
Distance and displacement: (units = m)
Distance is the path in which an object moves (scalar quantity).
Displacement is the shortest direct distance from start to end (vector quantity).
2.1.2 Speed and velocity: (units = m/s)
Speed is how fast an object moves within a given distance.
Average speed is the rate at which a given distance is covered.
average speed = $\frac{dis\mathrm{tan}ce}{time}$
average speed = $\frac{initialspeed+finalspeed}{2}$
Velocity (symbol v) is the rate at which displacement is changed.
Speed is the magnitude of velocity (a scalar quantity). Velocity is a combination of speed and direction (a vector quantity).
Example 1:
A commuter bus operates between Gwanda and Bulawayo which are 125 km apart; it takes 2 hours 30 minutes to travel between the two towns. What is the average speed of the commuter bus between the two towns?
Answer:
Average speed = $\frac{dis\mathrm{tan}ce}{speed}$ = $\frac{120km}{2.5h}$ = 50 km/h
Acceleration: (units = m/s2)
This is the rate at which velocity changes.
Acceleration can be; how fast an object speeds up, slows down or changes direction.
A negative acceleration is called deceleration or retardation.
Deceleration can therefore occur when:
object gradually decreases velocity, or
Velocity is negative (opposite direction to the prescribed direction).
When an object moves at a constant (steady) acceleration, it is said to be having a uniform acceleration
average acceleration = $\frac{changeinvelocity}{timetaken}$
a = $\frac{v-u}{t}$
[Where: v is the final velocity; u is the initial velocity]
Example2.
The car above passes the first post with a velocity of 20 m/s. If it has a steady acceleration of 2 m/s2, what is the velocity 5 s later, at the second post?
Answer:
a = $\frac{v-u}{t}$ v = at + u = (2 $m/{s}^{2}$)(5 s) + 20 m/s = 30 m/s
Investigating acceleration using a tickertape timer:
Ticker tape timer makes 50 dots per seconds onto a paper tape that is being pulled through the timer.
The size of the gap between successive dots depends on the speed of the tape going through the timer.
If the dots are far apart, the speed of the tape through the time was fast.
If the dots are close together, the tape was moving slowly through the timer.
2.1.5 Analysing the ticker tape
A trolley running down a ramp has its motion studied using a ticker tape timer.
The setup is shown below:
The length of tape produced by the trolley as it runs down the ramp is shown below.
The complete tape is cut into lengths with equal number of dots on each length
The lengths 1, 2, 3, 4 and 5 represent equal time intervals.
The length can be stuck onto a grid paper as shown below.
The above shows an increase in speed of the trolley.
The speed and acceleration can be found from the tape.
The speed is calculated by dividing the length of the tape with the time taken. (In the above case, there are 5 dots per cut strip. This means that each cut interval is 0.1 s).
Time taken = $\frac{numberofdots}{50}$ = $\frac{5}{50}$ = 0.1 s
Motion equations
The motion of a uniformly accelerated body can be described by three equations of motion:
1) v = u + at 2) s = ut + $\frac{1}{2}a{t}^{2}$ 3) ${v}^{2}={u}^{2}$ + 2as
u =initial velocity in m/s; v = final velocity in m/s; a = uniform acceleration of object in m/${s}^{2}$ s = distance moved after time t; t = time taken in s
Example 3:
A ball is accelerated at 5m/s2 from an initial velocity of 4 m/s for a time of 5 seconds. What is the final velocity and distance moved?
Answer:
Final velocity v = u + at = 4 m/s + (5 m/${s}^{2}$ x 5 s) = 29 m/s
Distance moved s = $ut+\frac{1}{2}a{t}^{2}$ = (4 m/s x 5 s) + ($\frac{1}{2}$x 5 m/${s}^{2}$ x ${5}^{2}$${s}^{2}$) = 82.5 m