Question

The position of an object moving along the x axis is given by the equation \(x = 1 + t^2\) (\(x\) in meter and \(t\) in second). At what time will the magnitude of its displacement and velocity be equal?

• A. \(t = 1 \text{ s}\)
• B. \(t = 1.5 \text{ s}\)
• C. \(t = 3 \text{ s}\)
• D. \(t = 2 \text{ s}\)

Hint:

Always distinguish between "position" (\(x\)) and "displacement" (\(\Delta x\)). Using \(x = v\) would lead to \(1 + t^2 = 2t \implies (t-1)^2 = 0 \implies t = 1\), which is a common distractor option.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
1. Displacement (\(\Delta x\)) is the change in position: \(\Delta x = x(t) - x(0)\).
2. Velocity (\(v\)) is the rate of change of position: \(v = dx/dt\).

Step 2: Key Formula or Approach:
Set \(|\Delta x| = |v|\) and solve for \(t\).

Step 3: Detailed Explanation:
1. Calculate position at \(t=0\):
\[ x(0) = 1 + 0^2 = 1 \]
2. Displacement at any time \(t\):
\[ \Delta x = (1 + t^2) - 1 = t^2 \]
3. Find velocity:
\[ v = \frac{d}{dt}(1 + t^2) = 2t \]
4. Equating magnitude of displacement and velocity:
\[ t^2 = 2t \]
\[ t^2 - 2t = 0 \implies t(t - 2) = 0 \]
Since we are looking for a non-zero time of motion, \(t = 2 \text{ s}\).

Step 4: Final Answer:
The magnitude of displacement and velocity are equal at \(t = 2 \text{ s}\).