A man covered a distance of 50 miles on his first trip, on a later trip he traveled 300 miles while going 3…

To solve this problem, we need to compare the time taken for the first trip with the time taken for the second trip.

1. Analyze the Question

• Trip 1: Distance ($d_1$) = 50 miles. Let the speed be $v$. Time ($t_1$) = $\frac{d_1}{v} = \frac{50}{v}$.
• Trip 2: Distance ($d_2$) = 300 miles. Speed is 3 times as fast, so new speed = $3v$.
• Goal: Compare the new time ($t_2$) with the old time ($t_1$).

2. Explain the Concept

The fundamental formula for motion is:

We will calculate the time for both trips and find the ratio between them.

3. Show the Solution

• Step 1: Calculate the time for the first trip ($t_1$) $$t_1 = \frac{50}{v}$$

• Step 2: Calculate the time for the second trip ($t_2$) $$t_2 = \frac{300}{3v}$$Simplify the fraction:$$t_2 = \frac{100}{v}$$

• Step 3: Compare $t_2$ with $t_1$ We want to see how $t_2$ relates to $t_1$: $$t_2 = \frac{100}{v}$$Since$t_1 = \frac{50}{v}$, we can rewrite$t_2$as:$$t_2 = 2 \times \left(\frac{50}{v}\right)$$ $$t_2 = 2 \times t_1$$

The new time is twice the old time.

4. Identify the Answer

Comparing the result to the given options, the new time is twice as much as the old time. This matches option C.

5. Exam Tip

When dealing with "times as fast" or "times as much" problems, always write out the variables clearly. Don't let the large numbers (50 and 300) distract you; focus on the relationship between the variables. If the distance increases by a factor of 6 (from 50 to 300) and the speed increases by a factor of 3, the time must increase by a factor of $6 \div 3 = 2$.