Understanding Acceleration on a Curved Road

When a car travels on a curved road, it experiences acceleration even if its speed is constant. This is because acceleration is a vector quantity, and its direction changes as the car follows the curve. The acceleration can be broken down into two components:

• Normal (Centripetal) Acceleration ($\mathbf{a_n}$): This component is always directed towards the center of the curvature of the path. It is responsible for changing the direction of the velocity vector, causing the car to follow a curved path.
• Tangential Acceleration ($\mathbf{a_t}$): This component is always directed along the tangent to the path, in the direction of motion if the speed is increasing, and opposite to the direction of motion if the speed is decreasing. It is responsible for changing the magnitude of the velocity vector (i.e., the speed).

Calculating Normal and Tangential Components

The formulas for the normal and tangential components of acceleration are:

• Normal acceleration, $\mathbf{a_n} = \frac{v^2}{r}$
• Tangential acceleration, $\mathbf{a_t} = \frac{dv}{dt}$

Where:

• $v$ is the speed of the car.
• $r$ is the radius of the curved road.
• $\frac{dv}{dt}$ is the rate of change of speed with respect to time.

Applying the Formulas to the Given Problem

We are given the following information for the car traveling on the curved road:

• Radius of the curved road, $r = 300$ m.
• Speed of the car, $v = 15$ m/s.

The problem states that the car is traveling at a speed of 15 m/s. This implies that the magnitude of the velocity, i.e., the speed, is constant. Therefore, the rate of change of speed with respect to time is zero.

Let's calculate the normal acceleration:

$\mathbf{a_n} = \frac{v^2}{r}$

$\mathbf{a_n} = \frac{(15 \text{ m/s})^2}{300 \text{ m}}$

$\mathbf{a_n} = \frac{225 \text{ m}^2/\text{s}^2}{300 \text{ m}}$

$\mathbf{a_n} = 0.75 \text{ m/s}^2$

Now, let's determine the tangential acceleration:

Since the speed of the car is constant ($v = 15$ m/s), the rate of change of speed is zero.

$\mathbf{a_t} = \frac{dv}{dt} = \frac{d}{dt}(15 \text{ m/s}) = 0 \text{ m/s}^2$

So, the normal and tangential components of acceleration are 0.75 m/s$^2$ and zero, respectively.

Conclusion

The normal component of acceleration is 0.75 m/s$^2$, and the tangential component of acceleration is zero. This matches the option stating 0.75 m/s$^2$ for the normal component and zero for the tangential component.