This doesn't sound logical

A car travels at a speed of 20 km/h for half the distance and at 30 km/h for the rest as shown in the diagram.

Find the average speed .

18 km/h 24 km/h 25 km/h 28 km/h 30 km/h Not enough information

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4 solutions

Syed Hamza Khalid
Sep 16, 2018

I have 2 ways of solving this problem:

(1) Using physics

Get this diagram:

• In this t 1 t_1 represents time taken from starting point to the center whereas t 2 t_2 represents time taken from center to the finishing point. Moreover x x determines the distance travelled.

Average speed = Total distance Total time = 2 x t 1 + t 2 > This is equation 1 \text{Average speed} = \dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{2x}{t_1 + t_2} \color{#20A900} ------------->\text{This is equation 1}

• Since it is uniform velocity over each sections of x x , therefore we can use s = u t s = ut because velocity is constant for x x but not for 2 x 2x .

s = u t x = 20 × t 1 > This is equation 2 x = 30 × t 2 > This is equation 3 \implies s = ut \\ x = 20 \times t_1 \color{#20A900} ------------->\text{This is equation 2} \\ x = 30 \times t_2 \color{#20A900} ------------->\text{This is equation 3}

• Now substitute equation 2 and 3 in equation 1 as follows:

Average speed = 2 x t 1 + t 2 = 2 x x 20 + x 30 = 2 x x ( 1 20 + 1 30 ) 2 ( 1 20 + 1 30 ) = 24 km/h \text{Average speed} = \dfrac{2x}{t_1 + t_2} \\ = \dfrac{2x}{\dfrac{x}{20} + \dfrac{x}{30}} \\ = \dfrac{2x}{x (\dfrac{1}{20} + \dfrac{1}{30})} \\ \implies \frac{2}{ (\dfrac{1}{20} + \dfrac{1}{30}) \\ \color{#69047E} = \boxed{\text{24 km/h}}}

(2)Using mathematics

Directly use the Harmonic mean formula:

f HM = k 1 a 1 + + 1 a k . f_{\text{HM}} = \frac{k}{\frac{1}{a_1} + \cdots + \frac{1}{a_k}}.

Or else in our case:

2 1 20 + 1 30 = 24 km/h \implies \frac{2}{ \dfrac{1}{20} + \dfrac{1}{30} \\ \color{#69047E} = \boxed{\text{24 km/h}}}


Some common errors are:

  1. You try to assume a distance. This is incorrect because time will change according to the distance.

  2. You try to use the arithmetic mean which will give you 25 km/h. It may be close but not accurate enough.

Assume the distance is 120 km, then the total time taken is 3 + 2 = 5 hours => the average is 120/5 = 24 km/h

Proof that this method works for any x:

Time taken over first half: x/20 h

Time taken over second half: x/30 h

Total time: x/20 + x/30 = 5x/60 h

Total average time: total distance / total time = 2x / (5x/60) = 120/5 = 24 km/h for ANY x

So you might as well take a convenient value for x, like 60 as I did.

Laszlo Mihaly
Sep 25, 2018

The average speed is v = d i s t a n c e t i m e v=\frac{distance}{time} . The time needed to take half the distance at 20km/h is d i s t a n c e / 2 20 k m / h \frac{distance/2}{20km/h} and for the other half of the trip d i s t a n c e / 2 30 k m / h \frac{distance/2}{30km/h} . The total time is t i m e = d i s t a n c e / 2 20 k m / h + d i s t a n c e / 2 20 k m / h time= \frac{distance/2}{20km/h}+\frac{distance/2}{20km/h} . Using this we get

v = d i s t a n c e t i m e = d i s t a n c e d i s t a n c e / 2 20 k m / h + d i s t a n c e / 2 20 k m / h = 2 1 20 + 1 20 = 24 k m / h v=\frac{distance}{time}= \frac{distance}{\frac{distance/2}{20km/h}+\frac{distance/2}{20km/h}} = \frac{2}{\frac{1}{20}+\frac{1}{20}}=24km/h

Chew-Seong Cheong
Sep 16, 2018

Let the total distance be D D . The time taken to travel the first half of the distance, t 1 = D 2 20 = D 40 h t_1 = \dfrac {\frac D2}{20} = \dfrac D{40} \text{ h} and that for the second half t 2 = D 2 30 = D 60 h t_2 = \dfrac {\frac D2}{30} = \dfrac D{60} \text{ h} . The average speed v a v = v_{av} = total distance traveled ÷ \div total time taken. v a v = D t 1 + t 2 = D D 40 + D 60 = 120 5 = 24 km/h \implies v_{av} = \dfrac D{t_1+t_2} = \dfrac D{\frac D{40} + \frac D{60}} = \dfrac {120}5 = \boxed{24} \text{ km/h} .

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