## Solution:

### Part a

We know that the formula for the distance traveled, d, if we are given rate, r, and time, t is given by $\displaystyle d=r\times t$

Therefore, the distance traveled is

$\displaystyle d=\left(4\:\frac{cm}{year}\times \frac{1\:year}{365\:\frac{1}{4}\:days}\times \frac{1\:day}{24\:hours}\times \frac{1\:hour}{60\:mins}\times \frac{1\:min}{60\:s}\right)\left(1\:s\right)$

$\displaystyle d=1.3\times 10^{-7}\:cm\:$

$\displaystyle d=1.3\times 10^{-9}\:m$

Therefore, the distance traveled by the tectonic plate is about $\displaystyle 1.3\times 10^{-9}\:m$          ☚

### Part b

We need to convert 4.0 cm/year to km/My.

$\displaystyle 4.0\:\frac{cm}{year}\times \frac{1\:m}{100\:cm}\times \frac{1\:km}{1000\:m}\times \frac{1\:000\:000\:years}{1\:My}=40\:\frac{km}{My}$

Therefore, the speed of the tectonic plates is 40 km/My.          ☚