Motion
Class 09 ScienceTo describe the position of an object you need to specify a reference point called the origin. Motion is a change of position. It can be described in terms of the distance moved or the displacement.
Motion Along a Straight Line
Consider the motion of an object moving along a straight path. The object starts its journey from O which is treated as its reference point. Let A, B and C represent the position of the object at different instants. At first, the object moves through C and B and reaches A. Then it moves back along the same path and reaches C through B.
The total path length covered by the object is OA + AC. This is the distance covered by the object. To describe distance you need to specify only the numerical value and not the direction of motion.
The shortest distance measured from the initial to the final position of an object is known as the displacement. For example, the distance of the final position C of the object from the initial position O.
Two different physical quantities - the distance and the displacement, are used to describe the overall motion of an object and to locate its final position with reference to its initial position at a given time.
Uniform and Non-Uniform Motion
Consider an object moving along a straight line. Let it travel 5 m in the first second, 5 m more in the next second, 5 m in the third second and 5 m in the fourth second. In this case, the object covers 5 m in each second. As the object covers equal distances in equal intervals of time, it is said to be in uniform motion.
Motions where objects cover unequal distances in equal intervals of time are called non-uniform motion.
Speed
Different objects may take different amounts of time to cover a given distance. Some of them move fast and some move slowly.
The distance travelled by the object in unit time is referred to as speed. The SI unit of speed is metre per second. This is represented by the symbol m s–1 or m/s. The other units of speed include centimetre per second (cm s–1) and kilometre per hour (km h–1). To specify the speed of an object, we require only its magnitude.
The speed of an object need not be constant. In most cases, objects will be in non-uniform motion. Therefore, we describe the rate of motion of such objects in terms of their average speed. The average speed of an object is obtained by dividing the total distance travelled by the total time taken.
$$ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} $$
If an object travels a distance s in time t then its speed v is,
$$ v = \frac{s}{t} $$
For example, a car travels a distance of 100 km in 2 h. Its average speed is 50 km h–1. The car might not have travelled at 50 km h–1 all the time. Sometimes it might have travelled faster and sometimes slower than this.
Example: An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object?
Total distance travelled by the object = 16 m + 16 m = 32 m
Total time taken = 4 s + 2 s = 6 s
$$ \text{Average speed} = \frac{32\,\text{m}}{6\,\text{s}} = \frac{16}{3}\,\text{m/s} = 5.33\,\text{m/s} $$
Velocity
The rate of motion of an object can be more comprehensive if we specify its direction of motion along with its speed. Velocity is the speed of an object moving in a definite direction. The velocity of an object can be uniform or variable. It can be changed by changing the object’s speed, direction of motion or both.
When an object is moving along a straight line at a variable speed, we can express the magnitude of its rate of motion in terms of average velocity. It is calculated in the same way as average speed.
In case the velocity of the object is changing at a uniform rate, then average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time.
$$ \text{Average velocity} = \frac{\text{Initial velocity} + \text{Final velocity}}{2} $$
$$ v_{\text{av}} = \frac{u + v}{2} $$
Example: The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h, calculate the average speed of the car in km h–1 and m s–1.
Distance covered by the car, s = 2400 km - 2000 km = 400 km
Time elapsed, t = 8 h
$$ \text{Average speed (in km/h)} = \frac{400\,\text{km}}{8\,\text{h}} = 50\,\text{km/h} $$
$$ \text{Average speed (in m/s)} = 50 \times \frac{1000}{3600} = 13.9\,\text{m/s} $$
Example: Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.
Total distance covered by Usha in 1 min is 180 m.
Displacement of Usha in 1 min = 0 m
$$ \text{Average speed} = \frac{180\,\text{m}}{60\,\text{s}} = 3\,\text{m/s} $$
$$ \text{Average velocity} = \frac{0\,\text{m}}{60\,\text{s}} = 0\,\text{m/s} $$
Acceleration
During uniform motion of an object along a straight line, the velocity remains constant with time. In this case, the change in velocity of the object for any time interval is zero. However, in non-uniform motion, velocity varies with time. It has different values at different instants and at different points of the path. Thus, the change in velocity of the object during any time interval is not zero.
Acceleration is a measure of the change in the velocity of an object per unit time.
$$ \text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time taken}} $$
If the velocity of an object changes from an initial value u to the final value v in time t, the acceleration a is,
$$ a = \frac{v - u}{t} $$
This kind of motion is known as accelerated motion. The acceleration is taken to be positive if it is in the direction of velocity and negative when it is opposite to the direction of velocity. The SI unit of acceleration is m s–2 .
Example: Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s–1 in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the next 5 s. Calculate the acceleration of the bicycle in both the cases.
In the first case:
initial velocity, u = 0; final velocity, v = 6 m s–1; time, t = 30 s
$$ a = \frac{6 - 0}{30} = \frac{6}{30} = 0.2 \, \text{m/s}^2 $$
In the second case:
initial velocity, u = 6 m s–1; final velocity, v = 4 m s–1; time, t = 5 s
$$ a = \frac{4 - 6}{5} = \frac{-2}{5} = -0.4 \, \text{m/s}^2 $$
Distance - Time Graphs
The change in the position of an object with time can be represented on the distance-time graph. In this graph, time is taken along the x-axis and distance is taken along the y-axis.
When an object travels equal distances in equal intervals of time, it moves with uniform speed. This shows that the distance travelled by the object is directly proportional to time taken. Thus, for uniform speed, a graph of distance travelled against time is a straight line.
Velocity - Time Graphs
The variation in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. In this graph, time is represented along the x-axis and the velocity is represented along the y-axis.
If the object moves at uniform velocity, the height of its velocity-time graph will not change with time. It will be a straight line parallel to the x-axis.
The product of velocity and time give displacement of an object moving with uniform velocity. The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of the displacement.
Equations of Motion
When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion.
$$ v = u + at $$
$$ s = ut + \frac{1}{2}at^2 $$
$$ v^2 = u^2 + 2as $$
where u is the initial velocity of the object which moves with uniform acceleration a for time t, v is the final velocity, and s is the distance travelled by the object in time t.
Example: A train starting from rest attains a velocity of 72 km h–1 in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.
Given,
u = 0
v = 72 km h–1 = 20 m s-1
t = 5 minutes = 300 s
(i) acceleration
$$ a = \frac{v - u}{t} = \frac{20 - 0}{300} = \frac{1}{15} \, \text{m/s}^2 $$
(ii) distance travelled
$$ s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times \frac{1}{15} \times 300^2 $$
$$ s = 3000 \, \text{m} = 3 \, \text{km} $$
Example: A car accelerates uniformly from 18 km h–1 to 36 km h–1 in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
Given
u = 18 km h–1 = 5 m s–1
v = 36 km h–1 = 10 m s–1
t = 5 s
(i) acceleration
$$ a = \frac{v - u}{t} = \frac{10 - 5}{5} = 1 \, \text{m/s}^2 $$
(ii) distance covered
$$ s = ut + \frac{1}{2}at^2 = 5 \times 5 + \frac{1}{2} \times 1 \times 5^2 $$
s = 25 m + 12.5 m = 37.5 m
Example: The brakes applied to a car produce an acceleration of 6 m s-2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.
Given
a = - 6 m s–2
t = 2 s
v = 0 m s–1
First, find initial velocity u
v = u + at
0 = u + (−6)(2)
u = 12 m/s
The, find distance s
$$ s = ut + \frac{1}{2}at^2 $$
$$ s = 12 \times 2 + \frac{1}{2} \times (-6) \times (2)^2 $$
= 24 m – 12 m = 12 m
Uniform Circular Motion
When the velocity of an object changes, we say that the object is accelerating. The change in the velocity could be due to change in its magnitude or the direction of the motion or both.
If the athlete moves with a velocity of constant magnitude along the circular path, the only change in his velocity is due to the change in the direction of motion. The motion of the athlete moving along a circular path is an example of an accelerated motion.
The circumference of a circle of radius r is given by 2πr. If the athlete takes t seconds to go once around the circular path of radius r, the speed v is given by
$$ v = \frac{2\pi r}{t} $$
When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.