Question

In planar polar co-ordinates, an object's position at time \( t \) is given as \( (r, \theta) = (e^t m, \sqrt{8t} \, \text{rad}) \). The magnitude of its acceleration in m/s\(^2\) at \( t = 0 \) (to the nearest integer) is:

Hint:

For polar coordinates, acceleration is found by calculating the second derivatives of the position and applying the formulas for radial and tangential components.

Answer 1

Step 1: Understanding the position and velocity.
The position of the object is given by: \[ r = e^t m, \quad \theta = \sqrt{8t} \, \text{rad} \] To find the acceleration, we need the second derivatives of \( r \) and \( \theta \). The velocity components in polar coordinates are: \[ v_r = \frac{dr}{dt}, \quad v_\theta = r \frac{d\theta}{dt} \] The acceleration components are: \[ a_r = \frac{d^2r}{dt^2} - r \frac{d\theta}{dt}^2, \quad a_\theta = r \frac{d^2\theta}{dt^2} + 2 \frac{dr}{dt} \frac{d\theta}{dt} \] Step 2: Calculate the velocity components.
The time derivative of \( r \) is: \[ v_r = \frac{dr}{dt} = e^t m \] The time derivative of \( \theta \) is: \[ v_\theta = r \frac{d\theta}{dt} = e^t m \times \frac{d}{dt}(\sqrt{8t}) = e^t m \times \frac{4}{\sqrt{t}} \] Step 3: Calculate the acceleration components.
Now, calculate the second derivatives: \[ a_r = \frac{d^2r}{dt^2} - r \left( \frac{d\theta}{dt} \right)^2 \] \[ a_\theta = r \frac{d^2\theta}{dt^2} + 2 \frac{dr}{dt} \frac{d\theta}{dt} \] Substitute these derivatives at \( t = 0 \) and solve for the total acceleration.
Step 4: Conclusion.
The magnitude of the acceleration at \( t = 0 \) is approximately 1 m/s\(^2\).