Question

A car goes from one end to the other end of a semicircular path of diameter ' \(\mathrm{d}\) '. Find the ratio between path length and displacement. \(\begin{array}{lll}\text { (A) }(3 \pi / 2) & \text { (B) } \pi \text { (C) } 2 & \text { (D) } \pi / 2\end{array}\)

Short answer

The ratio between the path length and displacement is \(\frac{\pi}{2}\).

Step-by-step solution

Calculate the path length

The car is traveling along a semicircular path with a diameter d. To find the path length or distance traveled by the car, we need to find half the circumference of the circle with diameter d. The formula for the circumference of a circle is given by: \[ C = \pi d\] Since the car is traveling along a semicircle, the path length (L) will be half the circumference: \[ L = \frac{1}{2} \pi d\]

Calculate the displacement

Displacement is the straight-line distance between the starting and ending points of the path. In this case, the starting and ending points lie on the diameter which is the straight line passing through the center of the semicircle. Therefore, the displacement (D) is equal to the diameter, d: \[ D = d\]

Find the ratio between path length and displacement

Now, we need to find the ratio between the path length (L) and the displacement (D). This can be calculated by dividing the path length by the displacement: \[ \text{Ratio} = \frac{L}{D}\] Substitute L and D with the values obtained from Step 1 and Step 2: \[ \text{Ratio} = \frac{\frac{1}{2} \pi d}{d}\]

Simplify the ratio

Now, we need to simplify the ratio obtained in Step 3. We can do so by canceling out the common factor 'd' from both numerator and denominator: \[ \text{Ratio} = \frac{\frac{1}{2} \pi d}{d} \times \frac{1}{d} = \frac{\frac{1}{2} \pi}{1} = \frac{\pi}{2}\] Thus, the ratio between the path length and displacement is \(\frac{\pi}{2}\). The correct answer is (D).

Key concepts

Semicircular Motion

When studying semicircular motion, consider how an object moves along a path that forms half of a circular arc. This type of motion is frequent in real-world scenarios, such as a car traversing a roundabout.
In semicircular motion:

• The path followed by the object is exactly half of the full circle.
• The starting and ending points are located at the ends of the diameter.
• It involves a shift from one side of the circle to the other, covering an arc length equal to half the circle's circumference.

It's crucial to note that while the path length refers to the actual distance traveled along the arc, displacement measures the direct straight-line distance from the initial to the final position.

Circumference of a Circle

The concept of the circumference is pivotal when dealing with circular and semicircular paths. The circumference, or the perimeter of a circle, measures the total distance around the circle. This concept is key when determining path lengths in problems involving circular or semicircular motion.
Here is how the circumference is determined:

• The formula for the circumference depends on the diameter or radius of the circle.
• When given the diameter, the formula is: \( C = \pi d \).
• When working with semicircular motion, only half of the circumference is considered: \( \frac{1}{2} \pi d \).

Understanding this formula assists in calculating distances along curved paths quickly and accurately.

Ratio Simplification

Simplifying ratios is a fundamental mathematical technique especially useful in comparing different measures. When you need to simplify the ratio between a path length and displacement, start by dividing those two values.
For example, simplifying the ratio of semicircular path length to displacement involves:

• Writing out the ratio with the actual numerical expressions in the fraction form.
• Using algebraic manipulation to cancel out any common factors.
• Performing arithmetic operations, if needed, to bring the ratio to the simplest form.

An example calculation: if the ratio is given as \( \frac{\frac{1}{2} \pi d}{d} \), simplification yields \( \frac{\pi}{2} \). Recognizing and applying this procedure aids in resolving more complex problems efficiently.