Speed, Distance & Time — Exercise 12(A)

Convert:

3.6 km/h to m/s

Answer

1 km = 1000 m and 1 h = 3600 s

$3.6 \text{ km/h } = \dfrac{3.6 \times 1000 \text{ m}}{3600 \text{ s}}\\[1em] = \dfrac{3600}{3600} \text{ m/s }\\[1em] = 1 \text{ m/s }$

Hence, 3.6 km/h = 1 m/s.

Convert:

54 km/h to cm/s

Answer

1 km = 1000 m = 1000 × 100 cm = 1,00,000 cm and 1 h = 3600 s

$54 \text{ km/h } = \dfrac{54 \times 1,00,000 \text{ cm}}{3600 \text{ s}}\\[1em] = \dfrac{54,00,000}{3600} \text{ cm/s }\\[1em] = 1500 \text{ cm/s }$

Hence, 54 km/h = 1500 cm/s.

Convert:

60 m/s to km/h

Answer

$1 \text{ m } = \dfrac{1}{1000} \text{ km and } 1 \text{ s } = \dfrac{1}{3600} \text{ h }\\[ 1em] 60 \text{ m/s }= \dfrac{60 \times \dfrac{1}{1000} \text{ km}}{\dfrac{1}{3600} \text{ h}}\\[1em] = \dfrac{60}{1000} \times 3600 \text{ km/h }\\[1em] = \dfrac{2,16,000}{1000} \text{ km/h }\\[1em] = 216 \text{ km/h }$

Hence, 60 m/s = 216 km/h.

Convert:

540 cm/s to km/h

Answer

$1 \text{ cm } = \dfrac{1}{1,00,000} \text{ km and } 1 \text{ s } = \dfrac{1}{3600} \text{ h }\\[1em] 540 \text{ cm/s } = \dfrac{540 \times \dfrac{1}{1,00,000} \text{ km}}{\dfrac{1}{3600} \text{ h}}\\[1em] = \dfrac{540}{1,00,000} \times 3600 \text{ km/h }\\[1em] = \dfrac{19,44,000}{1,00,000} \text{ km/h }\\[1em] = 19.44 \text{ km/h }$

Hence, 540 cm/s = 19.44 km/h.

Fill in the blanks:

Answer

(i) Distance = 129 m and time = 3 min

Speed = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{129 \text{ m}}{3 \text{ min}} = 43 \text{ m/min }$

Hence, speed = 43 m/min.

(ii) Speed = 40 km/h and time = 45 min

$45 \text{ min } = \dfrac{45}{60} \text{ h } = \dfrac{3}{4} \text{ h }$

Distance = Speed × Time = $40 \times \dfrac{3}{4} \text{ km } = 30 \text{ km }$

Hence, distance = 30 km.

(iii) Distance = 720 m and speed = 48 km/h

Converting the speed into m/s :

1 km = 1000 m and 1 h = 3600 s

$48 \text{ km/h } = \dfrac{48 \times 1000 \text{ m}}{3600 \text{ s}} = \dfrac{48000}{3600} \text{ m/s } = \dfrac{40}{3} \text{ m/s }\\[1em] \text{Time} = \dfrac{\text{Distance}}{\text{Speed}} = \dfrac{720 \text{ m}}{\dfrac{40}{3} \text{ m/s}}\\[1em] = 720 \times \dfrac{3}{40} \text{ s }\\[1em] = 54 \text{ s }$

Hence, time = 54 sec.

Rohit covers a distance of 140 m with the speed of 75 km/h. How much extra time will he take to cover a distance of 400 m with the same speed?

Answer

Extra distance to be covered = 400 m − 140 m = 260 m

Speed = 75 km/h

Converting the speed into m/s :

1 km = 1000 m and 1 h = 3600 s

$75 \text{ km/h } = \dfrac{75 \times 1000 \text{ m}}{3600 \text{ s}} = \dfrac{75000}{3600} \text{ m/s } = \dfrac{125}{6} \text{ m/s }\\[1em] \text{Extra time} = \dfrac{\text{Extra distance}}{\text{Speed}} = \dfrac{260 \text{ m}}{\dfrac{125}{6} \text{ m/s}}\\[1em] = 260 \times \dfrac{6}{125} \text{ s }\\[1em] = \dfrac{1560}{125} \text{ s }\\[1em] = 12.48 \text{ s }$

Hence, extra time taken = 12.48 sec.

Which is greater: 72 km/h or 21 m/s?

Answer

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

Converting 72 km/h into m/s :

1 km = 1000 m and 1 h = 3600 s

$72 \text{ km/h } = \dfrac{72 \times 1000 \text{ m}}{3600 \text{ s}} = \dfrac{72000}{3600} \text{ m/s } = 20 \text{ m/s }$

Since, 20 m/s < 21 m/s

⇒ 72 km/h < 21 m/s

Hence, 21 m/s is greater.

A journey is covered in 3 hours with a speed of 90 km h^-1. What must be the speed, if the same journey is to be completed in 5 hour?

Answer

Distance of the journey = Speed × Time = 90 km/h × 3 h = 270 km

New time = 5 h

New speed = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{270 \text{ km}}{5 \text{ h}} = 54 \text{ km/h }$

Hence, the required speed = 54 km/h.

A certain distance is covered in 50 minutes with a speed of 60 km per hour. How much extra time will be needed, if the distance is kept same and the speed is halved?

Answer

Speed = 60 km/h

Time = 50 min = $\dfrac{50}{60} \text{ h } = \dfrac{5}{6} \text{ h }$

Distance = Speed × Time = $60 \times \dfrac{5}{6} \text{ km } = 50 \text{ km }$

When the speed is halved,

$\text{New speed} = \dfrac{60}{2} \text{ km/h } = 30 \text{ km/h }\\[1em] \text{New time} = \dfrac{50 \text{ km}}{30 \text{ km/h}}\\[1em] = \dfrac{5}{3} \text{ h }\\[1em] = \dfrac{5}{3} \times 60 \text{ min }\\[1em] = 100 \text{ min }$

Extra time = 100 min − 50 min = 50 min

Hence, 50 minutes extra time will be needed.

Geeta covers 800 metres in 5 minutes.

(i) How much distance will she cover in half an hour?

(ii) How much time will she take to cover 6.4 km?

[Take the speed same for both the cases.]

Answer

Speed = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{800 \text{ m}}{5 \text{ min}} = 160 \text{ m/min }$

(i) Time = half an hour = 30 min

Distance = Speed × Time = 160 × 30 m = 4800 m = 4.8 km

Hence, she will cover 4800 m = 4.8 km.

(ii) Distance = 6.4 km = 6.4 × 1000 m = 6400 m

Time = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{6400 \text{ m}}{160 \text{ m/min}} = 40 \text{ min }$

Hence, she will take 40 minutes.

A bus, going at 60 km per hour, takes 4 hours to travel from Delhi to Agra.

(i) How much time will the bus take to cover the same distance, if its speed is increased by 20 km per hour?

(ii) How much extra time will the bus take in going from Delhi to Agra, if its speed is reduced by 20 km per hour?

Answer

Distance from Delhi to Agra = Speed × Time = 60 km/h × 4 h = 240 km

(i) Increased speed = 60 km/h + 20 km/h = 80 km/h

Time = $\dfrac{240 \text{ km}}{80 \text{ km/h}} = 3 \text{ h }$

Hence, the bus will take 3 hours.

(ii) Reduced speed = 60 km/h − 20 km/h = 40 km/h

Time = $\dfrac{240 \text{ km}}{40 \text{ km/h}} = 6 \text{ h }$

Extra time = 6 h − 4 h = 2 h

Hence, the bus will take 2 hours extra.

An animal is walking at a speed of 6 km per hour.

(i) How much distance will it walk in 8 minutes?

(ii) How much time will it take to cover 600 metres with the same speed?

Answer

(i) Speed = 6 km/h and time = 8 min

$8 \text{ min } = \dfrac{8}{60} \text{ h } = \dfrac{2}{15} \text{ h }\\[1em] \text{Distance} = \text{Speed} \times \text{Time} = 6 \times \dfrac{2}{15} \text{ km }\\[1em] = \dfrac{12}{15} \text{ km }\\[1em] = 0.8 \text{ km } = 800 \text{ m }$

Hence, it will walk 0.8 km = 800 m.

(ii) Distance = 600 m = $\dfrac{600}{1000} \text{ km } = 0.6 \text{ km }$

$\text{Time} = \dfrac{\text{Distance}}{\text{Speed}} = \dfrac{0.6 \text{ km}}{6 \text{ km/h}}\\[1em] = 0.1 \text{ h }\\[1em] = 0.1 \times 60 \text{ min } = 6 \text{ min }$

Hence, it will take 6 minutes.

Distance between two stations A and B is 80 km. Manoj goes from station A to station B with a certain speed and returns back to station A, with the same speed, in 5 hours. Find the speed of Manoj.

Answer

Total distance covered = 80 km + 80 km = 160 km

Total time taken = 5 h

Speed = $\dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{160 \text{ km}}{5 \text{ h}} = 32 \text{ km/h }$

Hence, the speed of Manoj = 32 km/h.

The initial distance between two cars A and B is 220 km. At 6 a.m., car A starts moving towards car B at 40 km/h and at 7 a.m. car B starts moving towards car A at 50 km/h. Find the time when cars A and B will meet.

Answer

Let car A meet car B after x hours.

Car A starts at 6 a.m., so car A moves for x hours.

Car B starts at 7 a.m., i.e. one hour later, so car B moves for (x − 1) hours.

Distance moved by car A = 40 × x = 40x km

Distance moved by car B = 50 × (x − 1) = 50(x − 1) km

When the cars meet,

Distance moved by car A + Distance moved by car B = 220 km

⇒ 40x + 50(x - 1) = 220

⇒ 40x + 50x - 50 = 220

⇒ 90x = 270

$\Rightarrow x = \dfrac{270}{90}$

⇒ x = 3

So, the cars meet 3 hours after 6 a.m.

Meeting time = (6 + 3) a.m. = 9 a.m.

Hence, cars A and B will meet at 9 a.m.

How much distance will be covered by a car in 24 seconds at 45 km/h?

Answer

Speed = 45 km/h

Converting the speed into m/s :

1 km = 1000 m and 1 h = 3600 s

$45 \text{ km/h } = \dfrac{45 \times 1000 \text{ m}}{3600 \text{ s}} = \dfrac{45000}{3600} \text{ m/s } = 12.5 \text{ m/s }$

Time = 24 s

Distance = Speed × Time = 12.5 × 24 m = 300 m

Hence, the distance covered = 300 m.

A car is moving with a speed of 72 km h^-1 and another car is moving with a speed of 16 ms^-1. If both the cars start from the same place, in the same direction and at the same time, find the difference of distances covered by both the cars in:

(i) 1 second

(ii) 15 seconds

Answer

Converting the speed of the first car into m/s :

1 km = 1000 m and 1 h = 3600 s

$72 \text{ km/h } = \dfrac{72 \times 1000 \text{ m}}{3600 \text{ s}} = \dfrac{72000}{3600} \text{ m/s } = 20 \text{ m/s }$

Speed of the second car = 16 m/s

As both the cars move in the same direction,

Relative speed = 20 m/s − 16 m/s = 4 m/s

(i) Difference of distances in 1 second = 4 m/s × 1 s = 4 m

Hence, the difference of distances in 1 second = 4 m.

(ii) Difference of distances in 15 seconds = 4 m/s × 15 s = 60 m

Hence, the difference of distances in 15 seconds = 60 m.

A distance of 6 km is covered in 8 minutes. If the same distance is to be covered in 4 minutes, by how much must the speed be increased?

Answer

Original speed = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{6 \text{ km}}{8 \text{ min}} = 0.75 \text{ km/min }$

New speed = $\dfrac{6 \text{ km}}{4 \text{ min}} = 1.5 \text{ km/min }$

Increase in speed = 1.5 km/min − 0.75 km/min = 0.75 km/min

Hence, the speed must be increased by 0.75 km/min.