Question

The position of a point in time \( t \) is given by \( x = \text{a} + \text{b}t - \text{c}t^2 \), \( y = \text{a}t + \text{b}t^2 \). Its resultant acceleration at time \( t \) in seconds is given by

• A. \( \text{b} - \text{c} \text{ unit} / \text{seconds}^2 \)
• B. \( \text{b} + \text{c} \text{ unit} / \text{seconds}^2 \)
• C. \( 2\text{b} - 2\text{c} \text{ unit} / \text{seconds}^2 \)
• D. \( 2\sqrt{\text{b}^2 + \text{c}^2} \text{ unit} / \text{seconds}^2 \)

Hint:

Acceleration is constant here because the position equations are quadratic in \(t\).

Answer 1

The Correct Option is D