Question

In planar polar co-ordinates, an object’s position at time \( t \) is given as \( (r, \theta) = (e^t \, \text{m}, \sqrt{8} \, t \, \text{rad}) \). The magnitude of its acceleration in m/s² at \( t = 0 \) (to the nearest integer) is:

Hint:

In polar coordinates, the total acceleration consists of both radial and tangential components. Don't forget to differentiate with respect to time to find these components.

Answer 1

Correct Answer: 9

Step 1: Understanding the position in polar coordinates.
The position of the object is given by \( r = e^t \) and \( \theta = \sqrt{8} \, t \). We need to find the acceleration of the object. Step 2: Finding velocity.
In polar coordinates, the velocity is given by the derivative of position with respect to time: \[ v_r = \frac{dr}{dt} \quad \text{and} \quad v_\theta = r \frac{d\theta}{dt} \] Using the given equations for \( r \) and \( \theta \), we find: \[ v_r = \frac{d}{dt}(e^t) = e^t \] \[ v_\theta = e^t \cdot \sqrt{8} \] Step 3: Finding acceleration.
The acceleration in polar coordinates is given by: \[ a_r = \frac{d}{dt}(v_r) \quad \text{and} \quad a_\theta = r \frac{d}{dt}(v_\theta) \] Calculating these, we get the total acceleration magnitude at \( t = 0 \) as 9 m/s².