LINEAR MOTION
Linear motion is movement in a straight line. Before we get to the main content of the topic, we first need to define some terms:
- Displacement – Distance in a defined direction.
- Speed – The distance covered per unit time / Rate of change of distance.
- Velocity – The displacement per unit time / Rate of change of displacement.
- Acceleration – The rate of change of velocity or speed.
Speed
Average speed is given by the expression: \text{Average speed} = \frac{\text{Distance covered}}{\text{Time taken}} \text{S} = \frac{\text{D}}{\text{T}} Speed is expressed in kilometres per hour (km/h) or metres per second (m/s). It is important to ensure that the appropriate units are used in the determination of speed.
Example
A minibus covered a distance of 180 km at an average speed of 90 km/h. It travelled at a speed of 80 km/h for \frac{2}{3} of its journey. At what speed did it travel the remaining part of the journey?
We must first find what distance the minibus covered while moving at 80 km/h. We are given that it was \frac{2}{3} of the journey, thus the distance in kilometres is \dfrac{2}{3}\times 180 =120 \text{ km}.
The time that it took to cover 120 km at 80 km/h is: \text{Time taken} = \frac{\text{Distance covered}}{\text{Average speed}} \text{T} = \frac{120}{80} = 1.5\text{ hours} The remaining distance after this is 180 – 120 = 60\text{ km}
The total time for the journey is: \text{T} = \frac{180}{90} = 2\ \text{hours} Thus, the time left is 2 – 1.5=0.5 \text{ hours} . To find the speed during the remaining part of the journey, we divide the remaining distance by the remaining time. \text{S} = \frac{\text{D}}{\text{T}} = \frac{60}{0.5} =\boxed{ 120\text{ km/h}}
To convert speed from kilometres per hour to metres per second, we multiply by \displaystyle \frac{5}{18} .
First, let us convert kilometres to metres and hours to seconds, individually. We know that 1 kilometre = 1000 metres and 1 hour = 3600 seconds.
Thus, 1 km/h can also be written as \displaystyle\frac{1000}{3600} m/s. We now simplify this fraction to get \displaystyle\frac{5}{18} m/s.
Now that we have the fraction, why do we multiply by it to convert km/h to m/s? In conversions, we are simply changing the units individually. When given a value in km/h, we first multiply it by 1000 (1 km = 1000 m). This is to change the numerator into metres.
Then, we need to divide the obtained value by 3600 (1 hour = 3600 seconds). This is done to change the unit of the denominator from hours to seconds.
In doing so, we are essentially just multiplying the initial value in km/h by \displaystyle\frac{1000}{3600} , which, as we saw earlier, simplifies to \displaystyle\frac{5}{18} . Therefore, we need to multiply by \displaystyle\frac{5}{18} to change speed from km/h to m/s.
On the other hand, to convert speed from metres per second to kilometres per hour, we multiply by \displaystyle \frac{18}{5} , the reciprocal of \displaystyle \frac{5}{18} .
Let us start with a certain quantity given in m/s. To convert meters into kilometers, the quantity is divided by 1000 and to convert seconds into hours, the quantity is divided by 3600. Thus, 1 m/s is equal to \cfrac{\cfrac{1}{1000}\text{ km}}{\cfrac{1}{3600}\text{ h}} . This is equal to \displaystyle\frac{3600}{1000} km/h. When simplified, we get \displaystyle\frac{18}{5} .
Example
A train moving at an average speed of 72 km/h takes 15 seconds to completely cross a bridge that is 80 metres long. Find the length of the train in metres.
We first convert convert the speed in km/h to m/s. 72\times\frac{5}{18}=20\text{ m/s} The distance covered by the train is given by: \text{D}=\text{S}\times\text{T} = 20\times15=300\text{ m}
Now, we must find a way to express the distance that the train moved. Let us do this by focusing on the front of the train. At the beginning of the bridge, the front of the train is just barely touching the bridge. After moving 80 metres, the front of the train is now barely touching the end of the bridge. The train now has to move through its length in order to completely cross the bridge. Taking the length of the train to be L: 80 + L = 300 L = 220\text{ m}
Acceleration
Acceleration is basically how quickly velocity changes. It is calculated as: \begin{align*}\text{Acceleration} &= \frac{\text{Change in velocity, m/s}}{\text{Change in time, s}} \\[12 pt] &=\frac{\text{Final velocity – Initial velocity}}{\text{Final time – Initial time}}\end{align*} It is expressed in metres per square second \left( \text{m}/\text{s}^2 \right).
Relative Speed
This is the main concept in this topic, so try to focus here. Relative speed is the speed of a moving body with respect to another.
When 2 bodies are moving in the same direction, their relative speed is the difference of their speeds.
When 2 bodies are moving in opposite directions, their relative speed is the sum of their speeds.
For example, a person moves at a speed of 7 m/s, and a second person moves at 9 m/s in the same direction. Then the second person will appear to be moving away at 2 m/s from the first person. This is because while the first person moves 7m forward in one second, the second one moves 9m forward. Thus, they get 2m farther apart each second.
If both people move towards each other, then they appear to be moving towards each other at 16 m/s. This is because in one second, the first person moves 7m towards the other, while the second person moves 9m towards the first. Therefore, in 1 second they get 16m closer.
Example 1
Two towns R and S are 245 km apart. A bus travelling at an average speed of 60 km/h left town R for town S at 8.00 am. A truck left town S for town R at 9.00 am and met with the bus at 11.00 am. Determine the average speed of the truck.
Generally when dealing with questions that involve two vehicles starting a journey at different times, we solve them by dividing the question into two parts. We first determine how far the first vehicle moved before the other vehicle started the journey, and subtract this distance from the original distance between them. Then, we calculate their relative speed to solve the rest of the question.
In this question, we start by finding the distance the bus moved before the truck left. It left at 8.00 am while the truck left at 9.00 am, thus the truck left 1 hour after the bus. The distance covered in this time is: \text{D}=\text{S}\times\text{T} = 60 \times 1 = 60 \text{ km} We now subtract this from the total distance to determine the distance between them at the time the truck left. 245-60 = 185\text{ km} We are told that the truck left at 9.00 am and met with the bus at 11.00 am, so they met 2 hours after the truck left. The vehicles covered a total distance of 185 km between each other in 2 hours. We can use this to find their relative speed. Since they are moving in opposite directions, the relative speed is the sum of their speeds. Taking the speed of the truck to be x : \text{Relative speed} = \frac{\text{Distance between vehicles}}{\text{Time taken}} 60+x=\frac{185}{2} 60+x=92.5 x=32.5\text{ km/h}
We can also take a different approach after finding the distance between the vehicles when the truck started its journey (185 km). We know that the truck met with the bus 2 hours after it left. In 2 hours, the bus moved a distance of 60\times2=120\text{ km} . We now subtract this distance from the distance between them at the time the truck left to get the distance that the truck covered. 185 – 120 = 65\text{ km} 65 km is the distance covered by the truck in 2 hours, thus we can calculate its speed. \text{S} = \frac{\text{D}}{\text{T}} = \frac{65}{2} = 32.5\text{ km/h}
Example 2
The distance between towns M and N is 280 km. A car and a lorry travel from M to N. The average speed of the lorry is 20 km/h less than that of the car. The lorry takes 1h 10 min more than the car to travel from M to N. If the speed of the lorry is x, find x.
The speed of the car is 20 km/h more than that of the lorry, thus it is equal to x+20 km/h. The time taken for the lorry to cover the distance is: \text{T} = \frac{\text{D}}{\text{S}}=\dfrac{280}{x}
Similarly, the time taken for the car to cover the distance is \dfrac{280}{x+20} . We are given that the difference in time is 1h 10 min, or \displaystyle\frac{7}{6} hours. \frac {280} {x} – \frac{280}{x+20}=\frac{7}{6}
Multiplying both sides of the equation by 6: \frac{1680}{x} – \frac{1680}{x+20}=7
Multiplying both sides of the equation by the LCM of the denominator, x(x+20) : 1680(x+20) – 1680x=7x(x+20) 1680x+33600 – 1680x=7x^2+140x 7x^2+140x +33600 = 0
Dividing by 7:x^2+20x +4800 = 0 (x+80)(x-60)=0 x = 60\text{ km/h}
Example 3
Two towns, A and B are 80 km apart. Juma started cycling from town A to B at 10.00 am at an average speed of 40 km/h. Malik started his journey from town B to A at 10.30 am and travelled by car at a speed of 80 km/h. Calculate the distance from town A at which they met and the time of day when the two met.
First, we find the distance that Juma covered before Malik started his journey. Juma started 30 minutes before Malik. \text{D}=\text{S}\times\text{T}=40\times\frac{30}{60}=20\text{ km} This means that when Malik departed, the distance between them was: 80 – 20 = 60\text{ km} .
Their relative speed is the sum of their individual speeds. RS = 40+80 = 120\text{ km/h}
To calculate the time it took to meet, we divide the distance between them by the relative speed. T = \frac{60}{120}=\frac{1}{2}\text{h}=30\text{ min}
The distance from town A is the sum of the distance covered by Juma before Malik started and the distance covered after Malik started his journey. \begin{align*}\text{Distance from A} &= 20 + \left( 40\times\frac{30}{60} \right)\\ &=20+20 \\ &= \boxed{40\text{ km}} \end{align*}
The time of day at which they met is equal to: 10.30\text{ am} + 30\text{ mins} = \boxed{11.00\text{ am}}