A bus is moving with a speed of `10 ms^-1` on a straight road. A scooterist wishes to overtake the bus in `100 s`. If the bus is at a distance of `1 km` from the scooterist with what speed should the scooterist chase the bus ?

Options

A. `20 ms^(-1)`
B. `40 ms^(-1)`
C. `25 ms^(-1)`
D. `10 ms^(-1)`

Answer

To solve the problem of how fast the scooterist must travel to overtake the bus, we can follow these steps:

### Step 1: Understand the Problem
The bus is moving at a speed of \(10 \, \text{ms}^{-1}\) and is \(1 \, \text{km}\) away from the scooterist. The scooterist wants to overtake the bus in \(100 \, \text{s}\).

### Step 2: Convert Distance to Meters
Since the distance is given in kilometers, we convert it to meters:
\[ 1 \, \text{km} = 1000 \, \text{m} \]

### Step 3: Identify the Variables
- Speed of the bus (\(V_b\)) = \(10 \, \text{ms}^{-1}\)
- Distance to be covered = \(1000 \, \text{m}\)
- Time to overtake = \(100 \, \text{s}\)
- Speed of the scooterist (\(V_s\)) = ?

### Step 4: Use the Concept of Relative Velocity
To overtake the bus, the scooterist must have a relative speed greater than the bus. The relative velocity (\(V_{rel}\)) can be expressed as:
\[ V_{rel} = V_s - V_b \]

### Step 5: Calculate the Required Relative Velocity
Using the formula for distance, we have:
\[ \text{Distance} = \text{Relative Speed} \times \text{Time} \]
Substituting the known values:
\[ 1000 \, \text{m} = (V_s - 10) \times 100 \]

### Step 6: Rearranging the Equation
Rearranging gives us:
\[ V_s - 10 = \frac{1000}{100} \]
\[ V_s - 10 = 10 \]

### Step 7: Solve for \(V_s\)
Now, add \(10\) to both sides:
\[ V_s = 10 + 10 \]
\[ V_s = 20 \, \text{ms}^{-1} \]

### Conclusion
The speed at which the scooterist must travel to overtake the bus in \(100 \, \text{s}\) is \(20 \, \text{ms}^{-1}\).