A bus is moving with a velocity `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 velocity should the scooterist chase the bus ?

Options

A. `50ms^(-1)`
B. `40ms^(-1)`
C. `30ms^(-1)`
D. `20ms^(-1)`

Answer

To solve the problem, we need to determine the velocity at which the scooterist must travel to overtake the bus in a given time. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
- The bus is moving with a velocity of \(10 \, \text{m/s}\).
- The distance between the scooterist and the bus is \(1 \, \text{km}\) (which is \(1000 \, \text{m}\)).
- The time in which the scooterist wants to overtake the bus is \(100 \, \text{s}\).

### Step 2: Set Up the Equation
Let the velocity of the scooterist be \(v \, \text{m/s}\). The effective velocity of the scooterist relative to the bus will be \(v - 10 \, \text{m/s}\) (since the bus is moving in the same direction).

### Step 3: Use the Distance Formula
We know the formula for distance:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
In this case, the distance to be covered by the scooterist to overtake the bus is \(1000 \, \text{m}\) in \(100 \, \text{s}\). Therefore, we can write:
\[ 1000 = (v - 10) \times 100 \]

### Step 4: Solve for \(v\)
Now, we can simplify the equation:
\[ 1000 = 100(v - 10) \]
Dividing both sides by \(100\):
\[ 10 = v - 10 \]
Now, add \(10\) to both sides:
\[ v = 20 \, \text{m/s} \]

### Conclusion
The scooterist must travel at a velocity of \(20 \, \text{m/s}\) to overtake the bus in \(100 \, \text{s}\).