Time Speed and Distance Quiz – 1

Man walked 6 km at 1.5 kmph, again he walked 8 km at speed of 2 kmph and 32 km at a speed of 8kmph
time taken individually:
=> 6/1.5 = 4 m
=> 8/2 = 4 m
=> 32/8 = 4 m
Average speed of man= total distance/ total time
=> 46/12 = 3 (5/6)

Let the speed of the trains be x and y respectively
length of train1 = 27x
length of train2 = 17y
Relative speed= x+ y
Time taken to cross each other = 23 s
=> (27x + 17 y)/(x + y) = 23
=> (27x + 17 y) = 23(x + y)
=> 4x = 6y
=> x/y = 6/4 = 3/2

Difference in timing = 15 – (–20) = 35 min.
Let required distance = x.
therefore, \frac{x}{10}-\frac{x}{15}=\frac{35}{60}
x = 17.5 km.

Speed = \frac{600}{5\times 60} m/sec = 2 m/sec
= 2 x \frac{18}{5}km/hr = 7.2 km/h

Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr
Distance covered in 6 minutes = (1/60) x 6 km = 1/10 km = 100 m
Therefore, Distance between the thief and policeman = (200 – 100) m = 100 m.

Car A travels at a speed of a and Car B travels at a speed of b.
Let us assume that they meet after t minutes.
Case I: Before they meet each-other.
Distance traveled by car A = a * t.
Distance traveled by car B = b * t.
Case II: After they cross each-other
Distance traveled by car A = a * 54.
Distance traveled by car B = 24 * b.
Distance traveled by car A after crossing car B = distance traveled by car B before crossing car A (and vice versa).
=> at = 54b ---------- (1)
and bt = 24a -------- (2)
Multiplying equations 1 and 2
we have ab * t^{2} = 54 * 24 * ab
=> t^{2} = 54 * 24
=> t = 36

Since we are required to find the time taken by B to travel the entire journey from city Q to P, therefore, B would take 36 + 24 = 60 minutes to complete his entire journey.

Car B and car C travel in opposite directions.
Their relative speed = Sum of the two speeds = 45 + 54 kmph = 99kmph. = 27.5 m/s
The relative distance = 220m. So, time they will take to cross each other \frac{220}{27.5} = 8 sec
Now, car A has to overtake car B within 8 seconds. The relative distance = 50m
Relative speed should be at least \frac{50}{8}= 6.25m/sec = 22.5km/hr
Car B travels at 45kmph, so car A should travel at at least 45 + 22.5 = 67.5kmph.

They meet for the first time 16 seconds after they start the race and for the second time 40 seconds from the time they start the race.
Now, we do not know their relative positions when they start the race. But we know that the time between the first and second meeting is 24 seconds. This is the time when they cover a relative distance of one lap length.
\frac{LapLength}{RelativeSpeed}= 24
Or Relative Speed = 5 m/s
The question says – Now, if B had started in the opposite direction to the one he had originally started, they would have met for the first time after 40 seconds.
Now, B would have crossed each other after 40 seconds if B had reversed direction. This is higher than the 24 seconds it takes them to cover the relative distance of a lap in the first instance.
Or, in the first instance they were travelling towards each other.
Or, a + b = 5 m/s.
They meet for the first time 16 seconds after they start the race.
Therefore, B’s speed = 3m/s

Distance covered by P in 8 sec = 4 * 8 = 32 m
Distance covered by P in 16 sec = 32 + 5 * 8 = 72 m
Distance covered by P in 24 sec = 72 + 6 * 8 = 120 m
Distance covered by P in 32 sec = 120 + 7 * 8 = 176 m
Distance covered by P in 40 sec = 176 + 8 * 8 = 240 m
Distance covered by P in 48 sec = 240 + 9 * 8 = 312 m
Distance covered by P in 56 sec = 312 + 10 * 8 = 392 m
Total distance in 56 sec = 392m
To cover balance 436 - 392 = 44 m with speed = 11 m/sec, time taken = 44/11 = 4 sec
So P would finish 436 m in 56 + 4 = 60 seconds.
In 60 seconds, Q cycles 60 * 6.5 = 390 m or B should have a 436 - 390 = 46m

During the journey, Average Speed = 2 × 400 × 600 / (400 + 600)= 480 km /hr.