Question

Cyclists A and B rode along a straight one way road starting from the same location and they rode for one hour. Cyclist A rode half the time with speed 10miles/hour and the other half time with speed 30 miles/hour. Cyclist B rode half the total distance with speed 10 miles/hour and half the distance with speed 30/hours.

Calculate the total distance traveled by cyclist B. Calculate the average velocity for cyclist B.

Answer 1

Well, the tricky thing in here is that for Cyclist B rode half the distance(!!) in one speed and the other half with another. That's unlike Cyclist A that rode half the time(!) in one speed and other half in another.
First let's define variables to the time Cyclist B rode at each speed
\[
x = \text{time_in_1st_speed} \\
y = \text{time_in_2nd_speed}
\]
Now since we know that
\[
\text{1st_speed} = 10_{miles/hour} \\
\text{2nd_speed} = 30_{miles/hour}
\]
We can say by the formula of
\[\text{Distance} = \text{Speed} \cdot \text{Time} \]
Since we know by the question that Cyclist B rode the same distance in both speeds:
\[
\text{distance_in_1st_speed} = \text{1st_speed} \cdot \text{time_in_1st_speed} \\
\text{distance_in_1st_speed} = 10_{miles/hour}\cdot x_{hours} = 10x_{miles} \\ \;\\
\text{distance_in_2nd_speed} = \text{2nd_speed} \cdot \text{time_in_2nd_speed} \\
\text{distance_in_2nd_speed} = 30_{miles/hour}\cdot y_{hours} = 30y_{miles} \\ \;\\
\text{distance_in_1st_speed} = \text{distance_in_2nd_speed} \\
10x_{miles} = 3yx_{miles} \quad// \div 10_{miles}\\
x = 3y
\]
Now, since we know he rode for one hour, and we can say
\[
\text{total_time} = x + y = 1_{hour} \\
x = 3y \quad \implies \quad \text{total_time} = 3y + y = 4y = 1_{hour} \;// \div 4\\
y = (\tfrac{1}{4})_{hour} \\
\text{distance_in_2nd_speed} = 30_{miles/hour} \cdot y_{hours} = \\
= \frac{30_{miles}}{4} = 7.5_{miles}
\]
And since that's half the distance:
\[
\text{total_distance} = 7.5_{miles} \cdot 2 = 15_{miles}
\]
I'll write the second part later, not enough time :|