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 = SB and Speed of Stream = Ss
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 = [V1 + V2]
- 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 = [V1 – V2]
- 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/