In everyday life, we see some objects at rest and others in motion. Birds fly, fish swim, blood flows through veins, and cars move. An object is said to be in motion when its position changes with time.
To describe the position of an object, we need to specify a reference point called the origin. For example, if we say a school is 2 km north of the railway station, the railway station is the reference point. Therefore, motion is a relative term—it depends on the observer and the reference point.
The simplest type of motion is motion along a straight line. Consider an object moving along a straight path. Let point $O$ be the origin.
Can displacement be zero? Yes! If the initial and final positions of an object are the same (like running one complete lap around a track), the displacement is zero, even though the distance covered is not zero.
Question: An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Solution:
Total time = $2 \text{ min } 20 \text{ s} = 140 \text{ s}$.
Number of rounds = $\frac{140}{40} = 3.5$ rounds.
Distance = $3.5 \times (\text{Circumference}) = 3.5 \times (2 \pi r) = 3.5 \times \pi \times 200 = 2200 \text{ m}$.
Displacement = After 3.5 rounds, the athlete is at the exact opposite end of the diameter. So, Displacement = Diameter = $200 \text{ m}$.
Different objects may take different amounts of time to cover a given distance. The rate of motion of an object is finding out the distance travelled by the object in unit time. This quantity is called speed.
The SI unit of speed is metre per second ($m/s$ or $m s^{-1}$). Other units include $cm/s$ and $km/h$. To specify speed, we only require its magnitude.
Since most objects in real life exhibit non-uniform motion, we describe their rate of motion in terms of average speed.
The rate of motion is more comprehensive if we specify its direction of motion along with its speed. The quantity that specifies both these aspects is called velocity. Velocity is the speed of an object moving in a definite direction.
Velocity can be changed by changing the object's speed, direction of motion, or both.
When an object moves along a straight line at a variable speed, we can express the magnitude of its rate of motion in terms of average velocity.
If velocity is changing at a uniform rate, then average velocity is the arithmetic mean of initial velocity ($u$) and final velocity ($v$):
The unit of velocity is also $m/s$.
Question: Abdul, while driving to school, computes the average speed for his trip to be 20 km/h. On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed for Abdul's trip?
Solution:
Let the distance to school be $x$.
Time for forward trip ($t_1$) = $\frac{x}{20}$ hours.
Time for return trip ($t_2$) = $\frac{x}{30}$ hours.
Total distance = $2x$.
Total time = $\frac{x}{20} + \frac{x}{30} = \frac{3x + 2x}{60} = \frac{5x}{60} = \frac{x}{12}$ hours.
Average speed = $\frac{\text{Total Distance}}{\text{Total Time}} = \frac{2x}{x/12} = 24 \text{ km/h}$.
In uniform motion along a straight line, velocity remains constant with time, meaning the change in velocity is zero. However, in non-uniform motion, velocity varies with time. This introduces a new physical quantity called acceleration.
Acceleration is a measure of the change in velocity of an object per unit time.
Where $u$ is initial velocity, $v$ is final velocity, and $t$ is time. The SI unit of acceleration is $m/s^2$ (metre per second squared).
Question: Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m/s in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m/s in the next 5 s. Calculate the acceleration of the bicycle in both cases.
Solution:
Case 1: Initial velocity $u = 0$, final velocity $v = 6 \text{ m/s}$, time $t = 30 \text{ s}$.
Case 2: Initial velocity $u = 6 \text{ m/s}$, final velocity $v = 4 \text{ m/s}$, time $t = 5 \text{ s}$.
Graphs provide a convenient method to present basic information about various events. For describing the motion of an object, we can use line graphs.
The change in position of an object with time can be represented on the distance-time graph. Time is taken along the x-axis and distance along the y-axis.
The variation in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. Time is on the x-axis and velocity on the y-axis.
The area enclosed by the velocity-time graph and the time axis is equal to the magnitude of the displacement (or distance) travelled by the object in a given time interval.
When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration, and distance covered in a certain time interval by a set of equations known as the equations of motion.
There are three such equations:
Derived from the slope of the v-t graph ($a = \frac{v-u}{t}$).
Derived by finding the area under the v-t graph, which is the sum of a rectangle ($ut$) and a triangle ($\frac{1}{2}at^2$).
Derived by finding the area under the v-t graph as a trapezium, eliminating $t$ using the first equation.
Where:
$u$ = initial velocity
$v$ = final velocity
$a$ = uniform acceleration
$t$ = time
$s$ = distance (displacement) covered in time $t$.
Question: A train starting from rest attains a velocity of 72 km/h 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.
Solution:
Initial velocity $u = 0$. Final velocity $v = 72 \text{ km/h} = 72 \times \frac{5}{18} = 20 \text{ m/s}$.
Time $t = 5 \text{ mins} = 300 \text{ s}$.
(i) Acceleration ($a$):
(ii) Distance ($s$):
Using $2as = v^2 - u^2$:
If you run along a rectangular track, you have to change your direction at the corners. The more sides a track has, the more frequently you must change direction. An infinite number of sides forms a circle.
When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.
In uniform circular motion, the speed is constant, but the velocity is constantly changing because the direction of motion is continuously changing at every point along the circle. Thus, uniform circular motion is an accelerated motion.
If the radius of the circular path is $r$, the circumference is $2\pi r$. If an object takes time $t$ to go around this circular path once, its speed $v$ is given by:
Examples of uniform circular motion include the motion of the moon around the earth, an artificial satellite in a circular orbit, or a cyclist on a circular track.
Question: An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Solution:
Radius $r = 42250 \text{ km}$.
Time $t = 24 \text{ h}$.
To convert to km/s: $11055.4 \div 3600 = 3.07 \text{ km/s}$.
Mastering these typical numericals will guarantee you can solve any exam question from this chapter.
Question: 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.
Solution:
Question: A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.
Solution:
Question: A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of -0.5 m/s². Find how far the train will go before it is brought to rest.
Solution:
Question: A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Solution:
