# SPEED, VELOCITY AND ACCELERATION

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 =

• 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?
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:
1. object gradually decreases velocity, or
2. 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 =

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?
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 =
$\frac{5}{50}$
= 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) + 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?
= 4 m/s + (5 m/${s}^{2}$ x 5 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 ${5}^{2}$ ${s}^{2}$)