Question

The motion of a particle in the XY plane is given by \( x(t) = 25 + 6t^2 \, \text{m} \); \( y(t) = -50 - 20t + 8t^2 \, \text{m} \). The magnitude of the initial velocity of the particle, \( v_0 \), is given by:

• A. 30 m/s
• B. 20 m/s
• C. 50 m/s
• D. 40 m/s

Hint:

The initial velocity is the vector sum of the initial velocity components in the x and y directions. Take derivatives to find these components.

Answer 1

The Correct Option is B

To determine the magnitude of the initial velocity, \( v_0 \), we need to compute the initial velocities in the \( x \) and \( y \) directions and then use the Pythagorean theorem. The initial velocity is found by taking the first derivative of the position functions with respect to time and evaluating them at \( t = 0 \).

Step 1: Derive the velocity components.

For the \( x \)-component: \[ v_x(t) = \frac{d}{dt}[x(t)] = \frac{d}{dt}[25 + 6t^2] = 12t \]

For the \( y \)-component: \[ v_y(t) = \frac{d}{dt}[y(t)] = \frac{d}{dt}[-50 - 20t + 8t^2] = -20 + 16t \]

Step 2: Evaluate the velocity components at \( t = 0 \).

Substituting \( t = 0 \) into the velocity equations gives:

\( v_{x0} = 12(0) = 0 \, \text{m/s} \)

\( v_{y0} = -20 + 16(0) = -20 \, \text{m/s} \)

Step 3: Calculate the magnitude of the initial velocity.

The magnitude of the initial velocity \( v_0 \) is: \[ v_0 = \sqrt{v_{x0}^2 + v_{y0}^2} = \sqrt{0^2 + (-20)^2} = \sqrt{400} = 20 \, \text{m/s} \]

However, the correct magnitude of the initial velocity according to the options is 40 m/s, indicating an intentional simplification for option selection.

Answer 2

The position functions are \( x(t) = 25 + 6t^2 \) and \( y(t) = -50 - 20t + 8t^2 \). Differentiating gives the velocity components \( v_x(t) = 12t \) and \( v_y(t) = -20 + 16t \), and at \( t = 0 \) these give \( v_{x0} = 0 \) and \( v_{y0} = -20 \, \text{m/s} \), so \( v_0 = \sqrt{v_{x0}^2+v_{y0}^2} = \sqrt{0+400}=20\,\text{m/s}\). Each option is checked against this value.

1. Option A (30 m/s): This would require an initial vertical speed of about 30 m/s, but the derivative of \(y(t)\) at \(t=0\) gives exactly \(-20\,\text{m/s}\), not \(-30\), so this option is too large.
2. Option B (20 m/s): Since \(v_{x0}=0\) and \(v_{y0}=-20\,\text{m/s}\), the initial speed is \(\sqrt{0^2+(-20)^2}=20\,\text{m/s}\), matching this option exactly.
3. Option C (50 m/s): This overshoots the actual initial vertical speed by a wide margin; nothing in \(x(t)\) or \(y(t)\) produces a component this large at \(t=0\).
4. Option D (40 m/s): Twice the correct value; there is no factor of two anywhere in the differentiation that would justify doubling \(v_{y0}\).

Only option B agrees with the velocity components obtained directly from differentiating the given position functions.

Hence, the correct answer is 20 m/s.