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

Concept: For a particle moving along a straight line under constant linear acceleration, we can determine its displacement and position using the standard kinematic equations of motion: \[ S = S_0 + v_0 t + \frac{1}{2}at^2 \] We must establish a consistent coordinate direction convention: we will define displacement to the right as positive and displacement to the left as negative.

Step 1: Identifying parameters with correct signs.
Let us list the values given in the problem statement and assign signs based on our direction convention:
• Initial reference position coordinate, \( S_0 = 0\text{ m} \)
• Initial velocity vector, \( v_0 = 12\text{ m/s} \) (directed to the right, so it is positive: \( +12 \))
• Acceleration rate, \( a = 2\text{ m/s}^2 \) (directed to the left, acting as a deceleration, so it is negative: \( -2 \))
• Elapsed travel time, \( t = 10\text{ s} \)

Step 2: Substituting values into the kinematic equation.
Substitute these parameters into the constant-acceleration position formula: \[ S = 0 + (12)(10) + \frac{1}{2}(-2)(10)^2 \]

Step 3: Simplifying individual terms.
Calculate the linear velocity displacement product term: \[ 12 \times 10 = 120\text{ m} \] Calculate the acceleration term product: \[ \frac{1}{2} \times (-2) \times (10)^2 = -1 \times 100 = -100\text{ m} \]

Step 4: Finding the final net position coordinate.
Combine the calculated terms to find the final position \( S \): \[ S = 120 - 100 = 20\text{ m} \] The final calculated position of the particle at time \( t=10\text{ s} \) is exactly \( 20\text{ m} \). This matches option (C).