DEMO

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. Fig.2.1.1.jpg (106 KB)

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 = distancetime

average speed = initial speed + final speed2

  • 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 = distancespeed
  = 120km2.5h
  = 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 = change in velocitytime taken

a = v-ut


 [Where: v is the final velocity; u is the initial velocity]
Example2.
5.jpg (84 KB)

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 = v-ut
v = at + u
= (2 m/s2)(5 s) + 20 m/s
= 30 m/s

Investigating acceleration using a tickertape timer:

6. Fig.2.1.2.jpg (81 KB)

  • 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.

7.jpg (76 KB)

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:

8. Fig. 2.1.3.jpg (97 KB)

  • The length of tape produced by the trolley as it runs down the ramp is shown below.

9.jpg (58 KB)

  • The complete tape is cut into lengths with equal number of dots on each length

10.jpg (65 KB)

  • The lengths 1, 2, 3, 4 and 5 represent equal time intervals.
  • The length can be stuck onto a grid paper as shown below.

11.jpg (117 KB)

  • 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 = number of dots50 
550 
= 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 + 12at2
3) v2 = u2 + 2as

u = initial velocity in m/s; v = final velocity in m/s;
                  a = uniform acceleration of object in m/s2
                  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/s2 x 5 s)
= 29 m/s

Distance moved s = ut+12at2
= (4 m/s x 5 s) + (12x 5 m/s2 x 52 s2)
= 82.5 m