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SPEED WITH DIRECTION
SPEED WITH DIRECTION
Figure 8.2(a):
Figure 8.2(b):
Look at the situations given in Fig. 8.2. If the bowling speed is 143 kmh–1 in Fig. 8.2(a) what does it mean? What do you understand from the signboard in Fig. 8.2(b)?
Different objects may take different amounts of time to cover a given distance. Some of them move fast and some move slowly. The rate at which objects move can
be different. Also, different objects can move at the same rate. One of the ways of measuring the rate of motion of an object is to find out the distance travelled by the object in unit time. This quantity is referred to as speed. The SI unit of speed is metre per second. This is represented by the symbol ms–1 or m/s. The other units of speed include centimetre per second (cm.s–1) and kilometre per hour (kmh–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. That is,
average speed =\(\frac {total \,\, distance\,\, travelled }{total \,\,time \,\,taken}\)
If an object travels a distance s in time t then its speed v is
v = \(\frac {s }{t}\) ______(8.1)
Let us understand this by an example. A car travels a distance of 100 km in 2 h. Its average speed is 50 kmh–1. The car might not have travelled at 50 kmh–1 all the time. Sometimes it might have travelled faster and sometimes slower than this.
The rate of motion of an object can be 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. 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 we calculate 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. That is
average velocity =\(\frac {intial \,\,velocity+final\,\,velocity }{2}\)
Mathematically, vav = \( \frac{u+v}{2}\) ______(8.2)
where vav is the average velocity, u is the initial velocity and v is the final velocity of the object. Speed and velocity have the same units that is, ms–1 or m/s.
Source: This topic is taken from NCERT TEXTBOOK
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Speed with direction - Activities and Illustrations
Activity 8.6:
* Measure the time it takes you to walk from your house to your bus stop or the school.
* If you consider that your average walking speed is 4 kmh–1, estimate the distance of the bus stop or school from your house.
Activity 8.7:
* At a time when it is cloudy, there may be frequent thunder and lightning. The sound of thunder takes some time to reach you after you see the lightning.
* Can you answer why this happens?
* Measure this time interval using a digital wristwatch or a stopwatch
* Calculate the distance of the nearest point of lightning. (Speed of sound in air = 346 ms-1.)
Illustration 8.1:
An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object?
Sol:
Total distance travelled by the object = 16 m + 16 m = 32 m
Total time taken = 4 s + 2 s = 6 s
Average speed = \(\frac{total \,\,distance \,\,travelled}{total \,\,time\,\,taken}\)
= \(\frac{32 m}{6s}\) = 5.33 ms-1
Therefore, the average speed of the object is 5.33 ms–1.
Illustration 8.2:
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.
Sol:
Distance covered by the car,
s = 2400 km – 2000 km = 400 km
Time elapsed, t = 8 h
Average speed of the car is,
vav = \(\frac{total \,\,distance \,\,travelled}{total \,\,time\,\,taken}=\frac{400km}{8h}=50kmh^{-1}\)
= 50 kmh–1
= \(=50\frac{km}{h}\times\frac{1000m}{1km}\times\frac{1h}{3600s}\)
= 13.9 m s–1
The average speed of the car is 50 kmh–1 or 13.9 ms–1.
Illustration 8.3:
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.
Sol:
Total distance covered by Usha in 1 min is 180 m.
Displacement of Usha in 1 min = 0 m
Average speed = \(\frac{total \,\,distance \,\,travelled}{total \,\,time\,\,taken}\)
= \(=\frac{180m}{1min}=\frac{180m}{1min}\times\frac{1min}{60s}\)
= 3 ms-1
Average velocity = \(\frac{Displacement}{total \,\,time\,\,taken}\)
= \(=\frac{0m}{60s}\)
= 0 ms–1
The average speed of Usha is 3 ms–1 and her average velocity is 0 ms–1.
Questions
1. Distinguish between speed and velocity.
2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
3. What does the odometer of an automobile measure?
4. What does the path of an object look like when it is in uniform motion?
5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 \(\times\) 108 ms–1.
Source: This topic is taken from NCERT TEXTBOOK