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
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Acceleration is constant here because the position equations are quadratic in \(t\).
Step 1: Understanding the Concept:
The position of a particle in 2D space is given by its \( x \) and \( y \) coordinates as functions of time \( t \).
The velocity vector components are the first derivatives of the position coordinates with respect to time.
The acceleration vector components are the second derivatives of the position coordinates with respect to time.
The resultant acceleration is the magnitude of the acceleration vector. Step 2: Key Formula or Approach:
Velocity components: \( v_x = \frac{dx}{dt} \), \( v_y = \frac{dy}{dt} \).
Acceleration components: \( a_x = \frac{dv_x}{dt} = \frac{d^2x}{dt^2} \), \( a_y = \frac{dv_y}{dt} = \frac{d^2y}{dt^2} \).
Resultant acceleration: \( a = \sqrt{a_x^2 + a_y^2} \). Step 3: Detailed Explanation:
Given the position coordinates:
\[ x = a + bt - ct^2 \]
\[ y = at + bt^2 \]
First, we find the velocity components by differentiating with respect to \( t \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(a + bt - ct^2) = b - 2ct \]
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(at + bt^2) = a + 2bt \]
Next, we find the acceleration components by differentiating the velocity components with respect to \( t \):
\[ a_x = \frac{dv_x}{dt} = \frac{d}{dt}(b - 2ct) = -2c \]
\[ a_y = \frac{dv_y}{dt} = \frac{d}{dt}(a + 2bt) = 2b \]
Now, we calculate the magnitude of the resultant acceleration:
\[ a = \sqrt{a_x^2 + a_y^2} \]
\[ a = \sqrt{(-2c)^2 + (2b)^2} \]
\[ a = \sqrt{4c^2 + 4b^2} \]
\[ a = \sqrt{4(c^2 + b^2)} = 2\sqrt{b^2 + c^2} \]
Step 4: Final Answer:
The resultant acceleration is \( 2\sqrt{b^2 + c^2} \text{ unit} / \text{seconds}^2 \).