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

Step 1: Concept Acceleration is the second derivative of position with respect to time (\(a = \frac{d^2r}{dt^2}\)).

Step 2: Meaning We need to find the components of acceleration \(a_x\) and \(a_y\) by differentiating the given equations twice.

Step 3: Analysis For \(x\): \(\frac{dx}{dt} = b - 2ct\) \(\rightarrow\) \(a_x = \frac{d^2x}{dt^2} = -2c\). For \(y\): \(\frac{dy}{dt} = a + 2bt\) \(\rightarrow\) \(a_y = \frac{d^2y}{dt^2} = 2b\). Resultant acceleration \(a = \sqrt{a_x^2 + a_y^2} = \sqrt{(-2c)^2 + (2b)^2} = \sqrt{4c^2 + 4b^2}\).

Step 4: Conclusion Factoring out the 4 gives \(2\sqrt{b^2 + c^2}\). Final Answer: (D)