A pickup vehicle is moving with a speed of 15.00 m/s on a straight road. A scooterist wishes to overtake the pickup vehicle in 150.0s. If the pickup vehicle is at an initial distance of 1.500 km from the scooterist, with what constant speed should the scooterist chase the pickup vehicle ?

Answer

To solve the problem of the scooterist overtaking the pickup vehicle, we can follow these steps:

### Step 1: Understand the Given Information
- The speed of the pickup vehicle (V_p) = 15.00 m/s
- The initial distance between the scooterist and the pickup vehicle (D) = 1.500 km = 1500 m
- The time in which the scooterist wants to overtake the pickup vehicle (t) = 150.0 s

### Step 2: Convert the Distance to Meters
Since the distance is given in kilometers, we convert it to meters for consistency with the speed units:
\[ D = 1.500 \, \text{km} = 1500 \, \text{m} \]

### Step 3: Use the Relative Velocity Concept
Let the speed of the scooterist be \( V_s \) (in m/s). The relative speed of the scooterist with respect to the pickup vehicle is:
\[ V_{relative} = V_s - V_p \]
Since the scooterist is trying to catch up with the pickup vehicle, we need to cover the distance \( D \) in time \( t \).

### Step 4: Set Up the Equation
The distance covered by the scooterist in time \( t \) can be expressed as:
\[ D = V_{relative} \times t \]
Substituting the expression for relative velocity:
\[ 1500 = (V_s - 15) \times 150 \]

### Step 5: Solve for \( V_s \)
Now, we can rearrange the equation to find \( V_s \):
\[ 1500 = 150V_s - 2250 \]
\[ 150V_s = 1500 + 2250 \]
\[ 150V_s = 3750 \]
\[ V_s = \frac{3750}{150} \]
\[ V_s = 25 \, \text{m/s} \]

### Conclusion
The constant speed at which the scooterist should chase the pickup vehicle is \( 25 \, \text{m/s} \).

---