Racecar Accident

Algebra Level 3

In a race, the average speed of the first lap was $200\text{ km/h}$ , but due to an accident the average speed of the second lap was $120\text{ km/h}$ . What is the average speed of the two laps?

150\text{ km/h}

160\text{ km/h}

170\text{ km/h}

180\text{ km/h}

4 solutions

Sabhrant Sachan
Jun 27, 2016

$\quad \text{Let the distance traveled in one lap be } "s" \\ \quad V_{\text{Avg.}} = \dfrac{\text{Total Distance}}{\text{Total Time taken}} \\ \quad V_{\text{Avg.}} = \dfrac{2\cancel{s}}{\frac{\cancel{s}}{200}+\frac{\cancel{s}}{120}} = \boxed{150 km/h}$

If you are finding the average you would add 200 and 120 and divide it by 2 to get 160.

Laura Earle - 4 years, 11 months ago

you are right, if the case was the time of the first lap equals the second ( $t_1 = t_2$ ), because $v_{avg} = \frac{d_{total}}{t_{total}} = \frac{d_1 + d_2}{t_1 + t_2} = \frac{d_1 + d_2}{2 \times t_1} = \frac{1}{2} ( \frac{d_1}{t_1} + \frac{d_2}{t_1} ) = \frac{1}{2} ( \frac{d_1}{t_1} + \frac{d_2}{t_2} ) = \frac{1}{2} (v_1 + v_2)$ , but as you see since the distance of the first lap equals to the distance of the second lap and they have different speeds so they must have different times which makes the statement $v_{avg} = \frac{v_1 + v_2}{2}$ invalid. and so the average speed is as @Sambhrant Sachan explained.

and generalized for $n$ laps $v_{avg} = \frac{n}{ \frac{1}{v_1} + \frac{1}{v_2} + \dots + \frac{1}{v_n} }$ .

Mehdi K. - 4 years, 11 months ago

Your solution, Sachan, shows two different formulas - one with 'time taken' as a denominator. However, we are not given time taken, not are we given the distance of one lap to calculate time taken per lap - it does not aid in answering the problem.

Gerald Hoskins - 4 years, 11 months ago

Uh Typo, it should be km/h.

Mehul Arora - 4 years, 11 months ago

Sherry Weaver
Jun 28, 2016

SAMBHRANT is correct. Let s=the distance of one lap as we are not given that value. The time to finish the first lap is distance/ speed = s/200. Likewise the second lap thus takes s/120. The total time for both laps is the sum of the two which is (3s+5s)/600 = 2s/150.

They travelled at the slower speed for a longer period of time. So the slower speed is weighted more heavily in the average than the higher speed.

Stephano Andreas - 4 years, 11 months ago

Jun Arro Estrella
Jun 30, 2016

Just use the harmonic mean formula. Let $a=$ initial speed $b=$ second speed

Harmonic mean tells us that the average velocity is:

$v_{ave}= \ (\frac{2ab}{a+b}$ )

Ken Gene Quah
Jul 10, 2016

Because it is talking about the average speed, we can set the distance of one lap to anything we like.

The most simple way is to set the distance to $\gcd(200,120) = 600$ km (as 200 and 120 are the speeds given). We use greatest common divisor because now it is easy to calculate [time = distance divided by speed] as our two speeds are a divisor of the distance, allowing us to work with integers.

Then, using the formula time = distance divided by speed, the first lap is completed in $\frac{600}{200} = 3$ hours while the second lap is completed in $\frac{600}{120} = 5$ hours.

This means both laps are completed in 8 hours, and the distance of 2 laps are $600 \times 2 = 1200$ km. Then the average speed is $\frac{1200}{8} = \boxed{150}$ km/h. (using the formula speed = distance divided by time)

*A race of 1200km is unlikely, but this can be any value, with the ratio of the time took for the first lap to the time took for the second lap being 3:5 which will not change.

Racecar Accident

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