# Placement Aptitude : Distance, Time and Speed – Formulas and Concepts

##### Distance covered per unit time is called Speed, $Speed = \frac{Distance}{Time} = \frac{D}{T}$

There are two obvious variations for the same formula:

- $Distance = Speed * Time$; and
- $Time = \frac{Distance}{Speed}$

Q: If Arif walks 2km in 30 minutes. What is his walking speed?

Sol: Always remember to pick a common unit. It’s not a limitation but best practice is to follow kilometre/ hour (km/ hr) or metre/ second (m/s)

distance travelled = 2km

time taken = 30 min = 0.5 hour

$Speed = \frac{Distance}{Time} = \frac{2}{0.5}$ = 4km/ hr

Q: If Arif walks at a speed of 1m/ s. Then how much distance will he cover in an hour?

Sol: We have to always match the units. Let’s convert time into seconds

T = 1 hour = 60 min = 3600s

Distance = Speed * Time = 1m/s * 3600s = 3600m = 3.6km

Q: If Arif walks 3km with a speed of 1m/s. Then how much time will he take to cover the distance?

Sol: We have to always match the units. Let’s convert distance into meters

Distance = 3 km = 3000m

Time = Distance / Speed = 3000m/ (1m/s) = 3000s = 50 min

**Conversion**

- To convert Speed in kmph to m/s — multiply by 5/18
- To convert Speed in m/s to kmph — multiply by 18/5

Q. A car travels from point A to point B at a speed of 60 km/h and returns from point B to point A at a speed of 40 km/h. What is the average speed for the entire journey?

Solution: Let’s assume the distance between point A and point B is D km

- Time to travel from A to B = Distance/Speed = D/60
- Time to travel from B to A = D/40

Total Distance=D+D=2D

The total time for the entire journey = D/60 + D/40 = 5D/120 = D/24

Average speed is calculated as = 2D/ (D/24) = 2×24 = 48 km/h

Q. A car covers the first part of a journey at a speed of 50 km/h and the second part at 70 km/h. The distance covered in the first part is twice the distance covered in the second part. What is the average speed for the entire journey?

Solution: Let the distance covered in the second part be D. So, the distance covered in the first part is 2D.

- Time for the first part (at 50 km/h) = $ \frac{2D}{50} $
- Time for the second part (at 70 km/h) = $ \frac{D}{70} $

The total distance is:Total Distance=2D+D=3D

The total time is = 2D/50 + D/70 = 19D/ 350

Average Speed = Total Distance / Total Time = 3D/(19D/350) = 55.26km/h

##### Problems on Trains

Trains are a special case of application of Time, Speed and Distance. Some common cases are discussed below.

**CASE 1 –** Train crossing a stationary object without length

$Time = \frac{L_t}{V_t} = \frac{Length \ of \ Train}{Speed \ of \ Train}$

**CASE 2 –** Train crossing a stationery object with length

$Time = \frac{L_t + L_o}{V_t} = \frac{Length \ of \ Train + Length \ of \ Object}{Speed \ of \ Train} $

Q. A train 180 meters long is moving at a speed of 72 km/h. How long will it take to pass a man standing on the platform?

Convert speed to m/s: 72 km/h=72×1000/3600=20 m/s

Time taken to pass the man (distance = 180 m): Time=Distance/ Speed=180/20=9 s

##### Boats and Streams

Boats and Streams are another special case of application of Time, Speed and Distance.

Let;s assume Speed of Boat in Still Water = S_{B} and Speed of Stream = S_{s}

**Up Stream** – When boat and stream travel in opposite directions (against each other)

$T_{UpStream} = \frac{D}{S_B -S_S}$

**Down Stream** – When boat and stream travel in same direction (with flow of water)

$T_{DownStream} = \frac{D}{S_B + S_S}$

If “u” is speed of boat in downstream and “v” is speed of boat is upstream then we have –

Speed of boat in still water $[S_B] = \frac{u+v}{2}$

Speed of water current $[S_S] = \frac{u-v}{2}$

Q. A boat can travel 20 km downstream in 2 hours and the same distance upstream in 4 hours. Find the speed of the boat in still water and the speed of the stream.

Sol: Downstream speed = Distance/ Time = 20/2=10 km/h

Speed upstream = 20/4 = 5 km/h

Speed of boat in still water = (Speed down stream + Speed up stream)/ 2 = (10+5)/2=7.5 km/h

Speed of the stream = (Speed down stream−Speed up stream) / 2= (10−5)/2=2.5 km/h

##### Circular Motion

A special case of movement is when two or more bodies are moving around a circular track.

**Opposite Directions** – When two persons are running around a circular track in opposite directions, then

- Relative Speed = [V
_{1}+ V_{2}] - From one meeting point to the next meeting point, the two of them cover a distance equal to the
**“**Length of track**“**together.

**Same Direction –** When two persons are running around a circular track in the same direction, then

- Relative Speed = [V
_{1}– V_{2}] - From one meeting point to the next meeting point, the faster person covers one complete round more than the slower person.

##### Clocks

HOUR HAND | MINUTE HAND | SECOND HAND |

$$12 \ hr \rightarrow 360^\circ $$ | $$60 \ min \rightarrow 360^\circ $$ | $$60 \ sec \rightarrow 360^\circ $$ |

$$1 \ hr \rightarrow 30^\circ $$ | $$1 \ min \rightarrow 6^\circ $$ | $$1 \ sec \rightarrow 6^\circ $$ |

You can refer following videos for more inspiration:

Refer Topic: “Simple and Compound Interest” https://www.learntheta.com/placement-aptitude-simple-compound-interest/

Read more about LearnTheta’s AI Practice Platform: https://www.learntheta.com/placement-aptitude/