Question

If a particle has the initial velocity of \( v_0 = 12\text{ m/s} \) to the right at \( S_0 = 0 \). Then what is the position when \( t = 10\text{ s} \) and \( a = 2\text{ m/s}^2 \) to the left?

• A. \( 10\text{ m} \)
• B. \( 15\text{ m} \)
• C. \( 20\text{ m} \)
• D. \( 25\text{ m} \)

Hint:

Always pay close attention to direction keywords like ``to the left'' or ``to the right'' in kinematics problems. Misinterpreting ``to the left'' as a positive acceleration value is a common mistake that changes the answer significantly (giving \( 120 + 100 = 220\text{ m} \)).

Answer 1

The Correct Option is C

Step 1: Split the motion into two phases.
The particle starts at 12 m/s to the right and decelerates at 2 m/s\(^2\) to the left. It first slows down, stops, then reverses direction. Time to stop: \( 0 = v_0 - at_1 \Rightarrow t_1 = \frac{12}{2} = 6\text{ s} \).

Step 2: Distance covered while slowing down.
Using \( v^2 = v_0^2 - 2as_1 \), at the stopping point \( v=0 \), so \( s_1 = \frac{v_0^2}{2a} = \frac{144}{4} = 36\text{ m} \) to the right.

Step 3: Motion in the remaining time.
The particle now starts from rest and accelerates to the left for the remaining \( t_2 = 10 - 6 = 4\text{ s} \). Distance covered: \( s_2 = \frac{1}{2}at_2^2 = \frac{1}{2}(2)(16) = 16\text{ m} \) to the left.

Step 4: Find the net position.
\[ S = s_1 - s_2 = 36 - 16 = 20\text{ m} \] \[ \boxed{S = 20\text{ m}} \]