Question:medium

An aeroplane is moving in a circular path with a speed of \(450\ \text{kmph}\). What is the change in velocity in half revolution?

Show Hint

In half revolution of circular motion, velocity reverses direction, so change in velocity is \(2v\).
  • \(0\ \text{kmph}\)
  • \(450\ \text{kmph}\)
  • \(250\ \text{kmph}\)
  • \(900\ \text{kmph}\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Question:
Velocity is a vector quantity, having both magnitude and direction.
In uniform circular motion, speed is constant, but velocity changes because its direction changes.
After half a revolution, the object is moving in the exactly opposite direction to its initial velocity.
Step 2: Key Formula or Approach:
Let the initial velocity be \(\vec{v}_1\).
After half a revolution, the velocity vector is \(\vec{v}_2\).
Since the path is circular and it's half a revolution, \(\vec{v}_2 = -\vec{v}_1\).
Change in velocity \(\Delta \vec{v} = \vec{v}_2 - \vec{v}_1\).
Step 3: Detailed Explanation:

Let the constant speed be \(v = 450 \text{ Kmph}\).

Suppose at the start, the aeroplane is at the bottom of the circle moving to the right. Its initial velocity vector is \(\vec{v}_i = v\hat{i}\).

After half a revolution, the aeroplane is at the top of the circle. At this point, it is moving to the left. Its final velocity vector is \(\vec{v}_f = -v\hat{i}\).

The change in velocity is given by the vector difference:
\[ \Delta \vec{v} = \vec{v}_f - \vec{v}_i \]
\[ \Delta \vec{v} = (-v\hat{i}) - (v\hat{i}) = -2v\hat{i} \]

The magnitude of the change in velocity is:
\[ |\Delta \vec{v}| = | -2v | = 2v \]

Substituting the value of \(v\):
\[ \text{Magnitude of change} = 2 \times 450 = 900 \text{ Kmph} \]

Step 4: Final Answer:
The change in velocity in half a revolution for circular motion is twice the speed, which is \(2 \times 450 = 900\) Kmph.
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