A car completes a 200 km journey at an average speed of x km/h. The car completes the return journey of 200 km at an average speed of (x + 10) km/h. Show that the difference betwee... A car completes a 200 km journey at an average speed of x km/h. The car completes the return journey of 200 km at an average speed of (x + 10) km/h. Show that the difference between the time taken for each of the two journeys is 2000 / x(x + 10) hours.
Understand the Problem
The question describes a scenario where a car travels 200 km at a speed of x km/h and returns the same distance at a speed of (x+10) km/h. The goal is to show or prove that the difference in time between the two legs of the journey is equal to 2000 / x(x+10) hours. I am going to assume we need to show how we got to the requested equation.
Answer
The difference in time is $\frac{2000}{x(x+10)}$.
Answer for screen readers
The difference between the time taken for each of the two journeys is $\frac{2000}{x(x+10)}$ hours.
Steps to Solve
- Calculate the time taken for the first journey
The time taken for the first journey is the distance divided by the speed. $$ t_1 = \frac{200}{x} $$
- Calculate the time taken for the return journey
The time taken for the return journey is the distance divided by the speed. $$ t_2 = \frac{200}{x+10} $$
- Calculate the difference in time
The difference in time is $t_1 - t_2$. $$ t_1 - t_2 = \frac{200}{x} - \frac{200}{x+10} $$
- Simplify the expression
To simplify, find a common denominator and combine the fractions. $$ \frac{200}{x} - \frac{200}{x+10} = \frac{200(x+10) - 200x}{x(x+10)} $$
- Expand and simplify the numerator
Expand the numerator: $$ \frac{200x + 2000 - 200x}{x(x+10)} $$ Simplify: $$ \frac{2000}{x(x+10)} $$
The difference between the time taken for each of the two journeys is $\frac{2000}{x(x+10)}$ hours.
More Information
This result tells us that the difference in travel times is inversely proportional to both the initial speed $x$ and the product of the initial speed and the increase in speed $x + 10$. As $x$ increases, the time difference decreases, which makes sense because at higher speeds, the 10 km/h difference becomes less significant.
Tips
A common mistake is to subtract the times in the wrong order, resulting in a negative value. While technically the negative would just indicate which journey was faster, it's important to subtract the slower time from the faster time to get a positive difference, reflecting the magnitude of the time difference. Another common mistake is in the algebraic manipulation when combining the fractions, specifically not distributing the negative sign correctly or making errors when finding the common denominator.
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