Tip #1: Use the formula Speed = Distance/ time while ensuring that units are same

Question: In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. Calculate the duration of the flight.

Solution:

Let the average duration of the flight be ‘t’ hours. Distance = 600 km.

Average Speed (km/ hr) = 600/ t

Average Speed – 200 = 600 / (t + 0.5)

(600/ t) – 200 = 600 / (t + 0.5)

(600 – 200 t) (t + 0.5) = 600 t

=> 2t^{2} + t - 3 = 0 =>t = 1 hr

Duration of the flight = 1.5 hrs

Question: Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 km/hr. But he will reach there at 12 noon if he travels at 15 km/hr. At what speed must he travel to reach A at 1 P.M.?

Solution:

According to the question,

(D / 10) – (D / 15) = 2 =>D = 60 km.

Time taken to reach A at 10 km/hr = 60 / 10 = 6 hours. (Starting time is 8 AM)

Therefore, we need to find the speed to cover 60 km should be in 5 hours.

Tip #2: If a person travels same distances with different speeds, then the average speed is not the arithmetic mean but the harmonic mean

If a person covers a distance d first at x km/hr and then covers the same distance d at y km/hr, then the average speed is:

= Total distance travelled/ Total time taken

= 2d/ (d/x + d/y)

= 2d/ [(yd + xd)/xy]

= 2dxy/[d(x+y)]

= 2xy/ x+y (Harmonic mean of x and y)

Question: A travels 25km at 50 km/hr and then 25km again with 70km/hr. What is A’s average speed during the whole journey?

Solution:

Average speed for the whole journey = (2x50x70) / (50+70) = 58.3km/hr

Tip #3: If A runs x times faster than B, A’s speed is actually 1+x the speed of B

Question: A runs 1⅔ times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?

Solution:

Speed of A, Sa = 5/3 x Sb

Let the distance of the course be ‘d’ meters

Time taken by A to cover distance ‘d’ = Time taken by B to cover distance ‘d-80’

d/[5/3 x Sb] = (d-80)/Sb

3d = 5d – 400

=> 2d = 640 => d = 200m

Question: A runs 1⅔ times faster than B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?

Solution:

Speed of A, Sa = (1 + 5/3) x Sb = 8/3 x Sb

Let the distance of the course be ‘d’ meters

Time taken by A to cover distance ‘d’ = Time taken by B to cover distance ‘d-80’

d/[8/3 x Sb] = (d-80)/Sb

3d = 8d – 640

=> 5d = 640 => d = 128m

[Note: Here, A 5/3 times faster than B, i.e., A’s speed = B’s speed + 5/3 times B’s speed = 8/3 times B’s speed.]

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Speed and distance

Speed Distance Time: Formula for time speed and distance, Speed time and distance problems, Speed Distance time questions with solutions