A certain light truck can go around a curve having a radius of `100 m` with a maximum speed of `16 m / s`. To have the same acceleration, at what maximum speed (in `m / s` ) can it go around a curve having a radius of `25 m ?`

Answer

To solve the problem, we need to find the maximum speed at which the light truck can go around a curve with a radius of 25 m, given that it can go around a curve with a radius of 100 m at a maximum speed of 16 m/s while maintaining the same centripetal acceleration.

### Step-by-Step Solution:

1. **Understanding Centripetal Acceleration**:
The centripetal acceleration (a_c) for an object moving in a circular path is given by the formula:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the speed of the object and \( r \) is the radius of the circular path.

2. **Calculate the Centripetal Acceleration for the First Curve**:
For the first curve with a radius \( r_1 = 100 \, m \) and maximum speed \( v_1 = 16 \, m/s \):
\[ a_{c1} = \frac{v_1^2}{r_1} = \frac{(16 \, m/s)^2}{100 \, m} \]
\[ a_{c1} = \frac{256 \, m^2/s^2}{100 \, m} = 2.56 \, m/s^2 \]

3. **Set Up the Equation for the Second Curve**:
Now, we want to find the maximum speed \( v_2 \) for the second curve with a radius \( r_2 = 25 \, m \) such that the centripetal acceleration remains the same:
\[ a_{c2} = \frac{v_2^2}{r_2} \]
Since we want \( a_{c2} = a_{c1} \):
\[ \frac{v_2^2}{25 \, m} = 2.56 \, m/s^2 \]

4. **Solve for \( v_2 \)**:
Rearranging the equation gives:
\[ v_2^2 = 2.56 \, m/s^2 \times 25 \, m \]
\[ v_2^2 = 64 \, m^2/s^2 \]
Taking the square root:
\[ v_2 = \sqrt{64 \, m^2/s^2} = 8 \, m/s \]

5. **Final Answer**:
The maximum speed at which the light truck can go around a curve with a radius of 25 m, while maintaining the same acceleration, is:
\[ \boxed{8 \, m/s} \]