An object that is moving at a constant rate is said to be in uniform motion and this chapter involves the calculation of the time and distance of the uniform motion object. Uniform motion problems may involve objects going in same direction, opposite directions and round trips.

- Eg: 90 km/hr = [latex]90 \times \frac{5}{18}[/latex] = 25 meters per sec

- Eg: 25 m/sec = [latex]25 \times \frac{18}{5}[/latex] = 90 km/hr.

- Eg: The ratio of speeds of two vehicles is 2 : 3. Then time taken ratio of the two vehicles to cover the same distance will be 3 : 2.

- Eg: The ratio of speeds of two vehicles is 2 : 3. If each vehicle is given one hour time, then the distance covered by the two vehicles will be in the ratio 2 : 3.

- Eg: A is twice as fast as B and each given 1 hour time. If A covers 20 miles of distance in
one hour, then B will cover 10 miles of distance in one hour.

- Eg: David travels at a speed of 56 miles per hour. If he reduces his speed in the ratio 7 : 6,
his new speed = [latex]\frac{6 \times 56}{7}[/latex] = 48 miles per hour.

- Eg: John travels 300 miles at the speed 45 miles per hour and travels another 300 miles at the speed of 55 miles per hour. Find the average speed for the whole journey. Average speed for the whole journey is [latex]\frac{2 \times 45 \times 55}{45 + 55}[/latex] = 55 miles per hour.

Let's see some examples for applying these rules:

- Distances needs to be covered to cross the platform = Sum of the lengths of the train and platform

∴ Distance traveled to cross the platform = 300 + 500 = 800 m

Time taken to cross the platform is = [latex]\frac{Distance}{Speed}[/latex] = [latex]\frac{800}{20}[/latex] = 40 sec.

Hence, time taken by the train to cross the platform is 40 seconds.

- The distance covered by the train to cross the pole is = Length of the train

We know, Distance = Speed × Time

∴ Length of the train = Speed × Time = 20 × 30 = 600 m. Hence, length of the train is 600 meters.

- When two trains are running in opposite direction, relative speed = 60 + 48 = 108 kmph = [latex]108 \times \frac{5}{18}[/latex]m/sec = 30 m/sec.

Sum of the lengths of the two trains is sum of the distances covered by the two trains in the above relative speed. Then, sum of the lengths of two trains is = Speed × Time = 30 × 15 = 450 m.

When two trains are running in the same direction, relative speed = 60 - 48 = 12 km/h = [latex]12 \times \frac{5}{18}[/latex] = [latex]\frac{10}{3}[/latex] m/sec.

When the two trains running in the same direction, a person in the faster train observes that he crossed the slower train in 36 seconds. The distance he covered in 36 seconds in the relative speed is equal to the length of the slower train.

∴ Length of the slower train = [latex]36 \times \frac{10}{3}[/latex] = 120 m

and, length of the faster train = 450 - 120 = 330 m

Hence, the length of the two trains are 330 m and 120 m.

- Relative speed of the train to man = 54 - 6 = 48 km/h = [latex]48 \times \frac{5}{8}[/latex] m/sec = [latex]\frac{40}{3}[/latex] m/sec

When the train passes the man, it covers the distance which is equal to its own length in the above relative speed. We have 12 seconds for the train to cross the man. So, the length of the train = Relative Speed × Time = [latex]\frac{40}{3} \times 12[/latex] = 160 m.

Speed of the train = 54 km/h = [latex]54 \times \frac{5}{18}[/latex] m/sec = 15 m/sec. When the train crosses the platform, it covers the distance which is equal to the sum of lengths of the train and platform. We have the train taking 20 seconds to cross the platform.

So, the sum of lengths of train and platform = Speed of the train × Time = 15 × 20 = 300 m

∴ Length of train + Length of platform = 300 ⇒ 160 + Length of platform = 300 ⇒ Length of platform = 300 - 160 = 140 m

Hence, the lengths of the train and platform are 160 m and 140 m, respectively.

- Distance covered = 20 km

Time taken = 4 hours

We know, Speed = [latex]\frac{Distance}{Time}[/latex] = [latex]\frac{20}{4}[/latex] km/hr

Therefore, speed = 5 km/hr

- Speed of car = [latex]\frac{Distance \ covered}{Time \ taken}[/latex] = [latex]\frac{450}{60}[/latex] m/sec = [latex]\frac{15}{2}[/latex]

= [latex]\frac{15}{2}[/latex] × [latex]\frac{18}{5}[/latex] km/hr

= 27 km/hr

Distance covered by train = 69 km

Time taken = 45 min = [latex]\frac{45}{60}[/latex] hr = [latex]\frac{3}{4}[/latex] hr

Therefore, speed of trains = [latex]\frac{69}{\frac{3}{4}}[/latex] km/hr

= [latex]\frac{69}{1}[/latex] × [latex]\frac{4}{3}[/latex] km/hr

= 92 km/hr

Therefore, ratio of their speed i.e., speed of car/speed of train = [latex]\frac{27}{92}[/latex] = 27 : 92

- Using the unitary method;

Time taken to cover 60 km = 90 minutes

Time taken to cover 1 km = [latex]\frac{90}{60}[/latex]90/60 minutes

Time taken to cover 100 km = [latex]\frac{90}{60} \times 100[/latex] = 150 minutes

Formula of Speed = [latex]\frac{Distance}{Time}[/latex]

= [latex]\frac{60}{\frac{3}{2}}[/latex] km/hr [given 1 hour 30 min = 1 [latex]\frac{30}{60}[/latex] = 1 [latex]\frac{1}{2}[/latex] hours = [latex]\frac{3}{2}[/latex] hours]

= [latex]\frac{60}{1}[/latex] × [latex]\frac{2}{3}[/latex] km/hr = 40 km/hr

Now, using the formula of Time = [latex]\frac{Distance}{Speed}[/latex] = [latex]\frac{100}{40}[/latex] km/hr = [latex]\frac{5}{2}[/latex]hours

= [latex]\frac{5}{2}[/latex] × 60 minutes, (Since 1 hour = 60 minutes)

= 150 minutes

- Using the unitary method;

70 km is covered in 1 hour.

1 km is covered in [latex]\frac{1}{70}[/latex] hours.

210 km is covered in [latex]\frac{1}{70}[/latex] × 210 hours = 3 hours

Given: speed = 70 km/hr, distance covered = 210 km

Time = [latex]\frac{Distance}{Speed}[/latex] = [latex]\frac{210}{70}[/latex] hours = 3 hours

- Using the unitary method;

In 60 minutes, distance covered = 80 km

In 1 minute, distance covered = [latex]\frac{80}{60}[/latex] km

Speed = 80 km/hr

Time = 36 minutes or [latex]\frac{36}{60}[/latex] hours

We know, formula of Distance = Speed x Time

= 80 × [latex]\frac{36}{60}[/latex]

= 48 km

Therefore, the train will travel 48 km in 36 minutes.

- Let the distance to school be 1 km, then time taken to cover 1 km at the rate of 7 [latex]\frac{1}{2}[/latex] km/hr

= [latex]\frac{Distance}{speed}[/latex] = [latex]\frac{1}{\frac{15}{2}}[/latex] = [latex]\frac{2}{15}[/latex] hr = [latex]\frac{2}{15}[/latex] × 60 minutes = 8 minutes

Time taken to cover 1 km at the rate of 10 km/hr

= [latex]\frac{Distance}{speed}[/latex] = [latex]\frac{1}{10}[/latex] hr = [latex]\frac{1}{10}[/latex] × 60 minutes = 6 minutes

Therefore, difference in time taken = (8 – 6) minutes = 2 minutes

But actual difference in time is 20 minutes

When the difference in time is 2 minutes, distance to school = 1 km

When the difference in time is 1 minute, distance to school = [latex]\frac{1}{2}[/latex] km

When the difference in time is 20 minutes, distance to school = [latex]\frac{1}{2}[/latex] × 20 km

Therefore, the distance to school is 10 km.

It gives the relationship between time (t), rate (r) and distance (d).

- Given that,

Rate = 4 km/hr

Time = 2 hrs 45 minutes

New Speed = 16.5 km/hr

Now, Distance = speed x time

Initial distance = 4 x [latex]\frac{11}{4}[/latex] = 11 km/hr

Then, Time = [latex]\frac{initial distance}{Speed}[/latex]

⇒ Time = [latex]\frac{11}{16.5}[/latex]

⇒ Time = 40 minutes

Therefore, by 40 minutes the man will cover the same distance.

- Given that,

Distance travelled by the person = 600 m

Time = 5 minutes

Now, Speed = [latex]\frac{Distance}{Time}[/latex]

Speed = [latex]\frac{600}{5 * 60}[/latex]

Speed = 2 m/sec

Convert m/sec in kmph, i.e.

⇒ Speed = 2 x [latex]\frac{18}{5}[/latex]

⇒ Speed = 7.2 km/hr

Therefore, Speed of the person = 7.2 km/hr

- Given,

Time of train is 10 minutes late

Let,

The new speed = [latex]\frac{4}{3}[/latex] of its usual speed

As, Speed = [latex]\frac{Distance}{Time}[/latex] ⇔ Time = [latex]\frac{Distance}{speed}[/latex]

⇒ Time = [latex]\frac{4}{3}[/latex] of its usual time

So, ([latex]\frac{4}{3}[/latex] of usual time) - (usual time)= 10 minutes

⇒ [latex]\frac{1}{3}[/latex] of usual time = 10

⇒ usual time = 10 x 3 = 30 minutes

Therefore, usual time to cover the journey = 30 minutes

- Let,

[latex]x[/latex] = 20 kmph

[latex]y[/latex] = 5 kmph

Given, time took for the whole journey = 5 hours 48 minutes

Now, Average speed = [latex]\frac{2xy}{x + y}[/latex]

⇒ Average speed = [latex]\frac{2 * 20 * 5}{20 + 5}[/latex]

⇒ Average speed = [latex]\frac{200}{25}[/latex]

⇒ Average speed = 8 kmph

So, Distance travelled in 5 hours 48 minutes i.e 5 x [latex]\frac{4}{5}[/latex] hours = 8 x [latex]\frac{29}{5}[/latex] = 46.4 km

Therefore, Distance between the post - office from the village = [latex]\frac{46.4}{2}[/latex] = 23.2 km

- Given that,

Distance covered by goods train in (6 + 4) = 10 hours

Distance covered by express train in 4 hours

Speed of express train = 90 kmph

Let,

The speed of the goods train be [latex]x[/latex] kmph

Now,

Distance = Speed x Time

Distance covered by goods train in 10 hours = Distance covered by express train in 4 hours

⇒ 10 [latex]x[/latex] = 4 x 90

⇒ [latex]x[/latex] = [latex]\frac{360}{10}[/latex]

⇒ [latex]x[/latex] = 36

Therefore, speed of goods train = 36 kmph