Measurement of Time and Motion

Calculate the speed of a car that travels 150 metres in 10 seconds. Express your answer in km/h.

Answer

Distance covered = 150 m

Time taken = 10 s

Speed = $\dfrac{\text{Total distance covered}}{\text{Total time taken}} = \dfrac{150 \text{ m}}{10 \text{ s}} = 15$ m/s

To express this in km/h, we use 1 m/s = 3.6 km/h.

Speed = 15 x 3.6 = 54 km/h

A runner completes 400 metres in 50 seconds. Another runner completes the same distance in 45 seconds. Who has a greater speed and by how much?

Answer

Speed of the first runner = $\dfrac{400 \text{ m}}{50 \text{ s}} = 8$ m/s

Speed of the second runner = $\dfrac{400 \text{ m}}{45 \text{ s}} = 8.89$ m/s (approximately)

The second runner has the greater speed, as he covers the same distance in less time.

Difference in speed = 8.89 - 8 = 0.89 m/s

So, the second runner is faster by about 0.89 m/s.

A train travels at a speed of 25 m/s and covers a distance of 360 km. How much time does it take?

Answer

Speed = 25 m/s

Distance covered = 360 km = 360 × 1000 m = 360,000 m

Time taken = $\dfrac{\text{Total distance covered}}{\text{Speed}} = \dfrac{360000 \text{ m}}{25 \text{ m/s}} = 14400$ s

Converting into hours: $\dfrac{14400 \text{ s}}{3600 \text{ s/h}} = 4$ h

So, the train takes 14,400 s, that is, 4 h.

A train travels 180 km in 3 h. Find its speed in:

1. km/h
2. m/s
3. What distance will it travel in 4 h if it maintains the same speed throughout the journey?

Answer

(i) Speed = $\dfrac{\text{Total distance covered}}{\text{Total time taken}} = \dfrac{180 \text{ km}}{3 \text{ h}} = 60$ km/h

(ii) To express this in m/s, we use 1 km/h = $\dfrac{5}{18}$ m/s.

Speed = $60 \times \dfrac{5}{18} = \dfrac{300}{18} = 16.67$ m/s (approximately)

(iii) Distance covered in 4 h = Speed × Time = 60 km/h × 4 h = 240 km

The fastest galloping horse can reach the speed of approximately 18 m/s. How does this compare to the speed of a train moving at 72 km/h?

Answer

To compare the two speeds, both must be in the same unit.

Let us convert the speed of the train into m/s.

Speed of the train = $72 \times \dfrac{5}{18} = 20$ m/s

Speed of the horse = 18 m/s

Since 20 m/s is greater than 18 m/s, the train is faster than the galloping horse by 20 − 18 = 2 m/s.

Distinguish between uniform and non-uniform motion using the example of a car moving on a straight highway with no traffic and a car moving in city traffic.

Answer

A car moving on a straight highway with no traffic is an example of uniform motion, while a car moving in city traffic is an example of non-uniform motion.

Data for an object covering distances in different intervals of time are given in the following table. If the object is in uniform motion, fill in the gaps in the table.

Answer

In uniform motion, an object covers equal distances in equal intervals of time. Here, the object covers 8 m in every 10 s (speed = $\dfrac{8 \text{ m}}{10 \text{ s}} = 0.8$ m/s). Using this, the gaps are filled as shown below.

A car covers 60 km in the first hour, 70 km in the second hour, and 50 km in the third hour. Is the motion uniform? Justify your answer. Find the average speed of the car.

Answer

The motion is not uniform. In uniform motion, an object covers equal distances in equal intervals of time. Here, the car covers 60 km, 70 km and 50 km in three equal intervals of one hour each. Since these distances are unequal, the motion is non-uniform.

Total distance covered = 60 + 70 + 50 = 180 km

Total time taken = 3 h

Average speed = $\dfrac{\text{Total distance covered}}{\text{Total time taken}} = \dfrac{180 \text{ km}}{3 \text{ h}} = 60$ km/h

Which type of motion is more common in daily life—uniform or non-uniform? Provide three examples from your experience to support your answer.

Answer

Non-uniform motion is more common in daily life. Uniform motion is an idealisation, as we seldom find objects moving with a constant speed over long distances or for long intervals of time. Three examples of non-uniform motion from daily life are:

1. A car moving in city traffic, which speeds up, slows down and stops at signals.
2. A bus starting from a bus stop and stopping again at the next stop.
3. A bicycle moving on a hilly road, going slower uphill and faster downhill.

Data for the motion of an object are given in the following table. State whether the speed of the object is uniform or non-uniform. Find the average speed.

Answer

The speed of the object is non-uniform. In each interval of 10 s, the object covers a different distance (6 m, 4 m, 6 m, 5 m, 8 m, 6 m, 7 m, 3 m, 10 m and 5 m). Since unequal distances are covered in equal intervals of time, the motion is non-uniform.

Total distance covered = 60 m

Total time taken = 100 s

Average speed = $\dfrac{\text{Total distance covered}}{\text{Total time taken}} = \dfrac{60 \text{ m}}{100 \text{ s}} = 0.6$ m/s

A vehicle moves along a straight line and covers a distance of 2 km. In the first 500 m, it moves with a speed of 10 m/s and in the next 500 m, it moves with a speed of 5 m/s. With what speed should it move the remaining distance so that the journey is complete in 200 s? What is the average speed of the vehicle for the entire journey?

Answer

Total distance = 2 km = 2000 m

Time taken for the first 500 m = $\dfrac{500 \text{ m}}{10 \text{ m/s}} = 50$ s

Time taken for the next 500 m = $\dfrac{500 \text{ m}}{5 \text{ m/s}} = 100$ s

Time used so far = 50 + 100 = 150 s

Remaining distance = 2000 − 500 − 500 = 1000 m

Remaining time = 200 − 150 = 50 s

Speed for the remaining distance = $\dfrac{1000 \text{ m}}{50 \text{ s}} = 20$ m/s

So, the vehicle should move the remaining 1000 m at a speed of 20 m/s.

Average speed for the entire journey = $\dfrac{\text{Total distance covered}}{\text{Total time taken}} = \dfrac{2000 \text{ m}}{200 \text{ s}} = 10$ m/s