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. 40 m/s
• C. 50 m/s
• D. 20 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

The initial velocity is the derivative of the position functions with respect to time. Thus, the velocity components are: \[ v_x(t) = \frac{d}{dt} \left( 25 + 6t^2 \right) = 12t \] \[ v_y(t) = \frac{d}{dt} \left( -50 - 20t + 8t^2 \right) = -20 + 16t \] At \( t = 0 \): \[ v_x(0) = 12 \times 0 = 0 \] \[ v_y(0) = -20 \] The magnitude of the initial velocity is: \[ v_0 = \sqrt{v_x(0)^2 + v_y(0)^2} = \sqrt{0^2 + (-20)^2} = 40 \, \text{m/s} \]