A golfer rides in a golf cart at an average speed of 3.10 m/s for 28.0 s. She then gets out of the cart and starts walking at an average speed of 1.30 m/s. For how long ( in seconds ) must she walk if her average speed for the entire trip, riding and walking, is 1.80 m/s ?

Options

A. 73 s
B. 31 s
C. 57 s
D. 44 s

Answer

To solve the problem step by step, we will use the concept of average speed and the relationships between distance, speed, and time.

### Step 1: Identify the given data
- Average speed while riding in the golf cart (V1) = 3.10 m/s
- Time spent riding (T1) = 28.0 s
- Average speed while walking (V2) = 1.30 m/s
- Average speed for the entire trip (V_avg) = 1.80 m/s

### Step 2: Calculate the distance covered while riding
Using the formula:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
The distance covered while riding (D1) can be calculated as:
\[ D1 = V1 \times T1 = 3.10 \, \text{m/s} \times 28.0 \, \text{s} = 86.8 \, \text{m} \]

### Step 3: Express the distance covered while walking
Let the time spent walking be T2. The distance covered while walking (D2) can be expressed as:
\[ D2 = V2 \times T2 = 1.30 \, \text{m/s} \times T2 \]

### Step 4: Write the equation for average speed
The average speed for the entire trip is given by:
\[ V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} \]
The total distance is:
\[ \text{Total Distance} = D1 + D2 = 86.8 \, \text{m} + 1.30 \, \text{m/s} \times T2 \]
The total time is:
\[ \text{Total Time} = T1 + T2 = 28.0 \, \text{s} + T2 \]

Substituting these into the average speed formula gives:
\[ 1.80 = \frac{86.8 + 1.30 \times T2}{28.0 + T2} \]

### Step 5: Cross-multiply to solve for T2
Cross-multiplying gives:
\[ 1.80 \times (28.0 + T2) = 86.8 + 1.30 \times T2 \]
Expanding this:
\[ 50.4 + 1.80 \times T2 = 86.8 + 1.30 \times T2 \]

### Step 6: Rearranging the equation
Rearranging the equation to isolate T2:
\[ 1.80 \times T2 - 1.30 \times T2 = 86.8 - 50.4 \]
This simplifies to:
\[ 0.50 \times T2 = 36.4 \]

### Step 7: Solve for T2
Dividing both sides by 0.50 gives:
\[ T2 = \frac{36.4}{0.50} = 72.8 \, \text{s} \]

### Conclusion
The time she must walk (T2) is **72.8 seconds**.

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