CAT 2025 Lesson : Time & Speed - Relative Speed

A and B are two cars, $200$ km apart, and moving towards each other at a speed of $30$ km/hr and $50$ km/hr respectively. What is the distance covered by Car A when it meets Car B?

Solution

Time taken $= \dfrac{200}{30 + 50} = 2.5$ hours

Distance covered by Car A $= 30 \times 2.5$ = 75 km

Alternatively

Note that for every $30$ km covered by Car A, $50$ km is covered by Car B. So, the ratio of the distances covered will be the ratio of their speeds.

Ratio of distances covered by Cars A and B $= 3 : 5$ and the total distance is $200$ km.

Distance covered by Car A $= \dfrac{3}{8} \times 200$ = 75 km

Answer: $75$ km

From a point P, Jaggu and Maggu start driving in opposite directions at $40$ km/hr and $60$ km/hr respectively. How far from each other are they after $150$ minutes?

Solution

As the two are moving from a point in opposite directions, they will never meet. However, as they are going in opposite directions, their speeds are added to get the relative speed.

Relative Speed $= 40 + 60 = 100$ km/hr

Time taken $= \dfrac{150}{60} = 2.5$ hours

Distance between them after $2.5$ hours $= 2.5 \times 100 = 250$ km

Answer: $250$ km

When X and Y are $100$ m apart, they start running towards each other at speeds of $3$ m/s and $2$ m/s respectively. How many seconds have they run for when they are $200$ m apart?

Solution

In this question, X and Y bridge the gap of 100m to meet each other and then keep running in opposite directions till they are 200m apart.

As they are running in opposite directions, relative speed $= 3 + 2 = 5$ m/s

Time taken to meet $= \dfrac{100}{5} = 20$ seconds

Additional time taken to be $200$m apart $= \dfrac{200}{5} = 40$ seconds

Total time taken $= 20 + 40 = 60$ seconds

Alternatively

Together, they bridge the gap of $100$m and then create a gap of $200$m.

Time taken $= \dfrac{100 + 200}{5} = 60$ seconds

Answer: $60$ seconds

Bhaskar starts running from point A at a speed of 15 m/s. After a minute, Charlie starts running from the same point A along the same path as Bhaskar, at a speed of 20 m/s. What is the distance covered by Bhaskar when they meet? How many minutes after Bhaskar starts running does he meet Charlie?

Solution

When Charlie starts the race, Bhaskar has already run for a minute and the catch-up distance for Charlie is $15 \times 60 = 900$m. And, Charlie is faster than Bhaskar by $20 - 15 = 5$ m/s, which is the catch-up speed.

$\therefore$ Time taken for the two to meet $= \dfrac{900}{5} = 180$ seconds $= 3$ minutes

Distance covered by Charlie to the meeting point $= 20$ m/s $\times 180$ s = $3600$m $= 3.6$ km

And, when they meet, Bhaskar has run for a total of $3 + 1 = 4$ minutes

Answer: $3.6$ km and $4$ minutes

At $2$ pm, a car travelling at $64$ km/hr overtakes a lorry travelling at $48$ km/hr in the same direction. Exactly $2$ hours later, the car goes past a bus travelling at $32$ km/hr in the opposite direction. At what time do the bus and the lorry meet?

Solution

The car is $64 - 48 = 16$ km/hr faster than the lorry.

$2$ hours after overtaking, i.e.at 4 pm, the car is $16 \times 2 = 32$ km ahead of the lorry, and meets the bus.

$\therefore$ The bus and lorry are $32$ km apart at $4$ pm and travelling in opposite direction at $48 + 32 = 80$ km/hr.

Time taken for bus and lorry to meet $= \dfrac{32}{80} = 0.4$ hours = 24 minutes

$\therefore$ The bus and lorry meet at $24$ minutes past $4$ pm, i.e. $4:24$.

Answer: $4:24$ pm

When a shooter, stationary at point A, fires the gun in the air at $4:00$pm, a car starts from point A at a constant speed. When the shooter fires the gun for the second time at $4:30$pm, the shot is heard by the driver of the car at $4:40$pm. What is the speed of the car (in km/hr)? [Sound travels at a speed of $330$ m/s]

Solution

Let the speed of car be $x$ m/s. At $4:30$pm, it has travelled for $30$ minutes $= 1800$ seconds.

Distance between A and car at $4:30$pm $= x \space \text{m/s} \times 1800 \space \text{s} = 1800x$

Relative speed $=$ Speed of sound $-$ Speed of Car $= (330 - x)$ m/s

As it takes $10$ more minutes or $600$ seconds for the car driver to hear the gun shot,

Time taken $= \dfrac{\text{Distance}}{\text{Relative Speed}} = \dfrac{1800x}{330 - x} = 600$

⇒ $1800x = 198000 - 600x$

⇒ $x = \dfrac{198000}{2400} = \dfrac{330}{4}$ m/s

Speed of car is required in km/hr $= \dfrac{330}{4} \times \dfrac{18}{5} = 33 \times 9$ = 297 km/hr

Alternatively (Recommended Method)

The sound of the second bullet travels for $10$ minutes and then meets the car.
The car has travelled for $40$ minutes from point A when the sound is heard.

Ratio of time taken by car to sound $= 40 : 10 = 4 : 1$
$\therefore$ Ratio of speeds of car to sound $= 1 : 4$

Let the speed of car be $x$ m/s. The speed of sound is given to be $330$ m/s.

$\dfrac{x}{330} = \dfrac{1}{4}$ ⇒ $x = \dfrac{330}{4} = 82.5$ m/s

Speed of car in in km/hr $= 82.5 \times \dfrac{18}{5}$ = 297 km/hr

Answer: $297$ km/hr