Question

There is a tree in the middle of your front yard and you take pictures of a squirrel as it runs around the tree. When you take the first picture, the squirrel is 12.0 m away from the tree in the direction 30.0◦ N of E. Twenty seconds later, you take a second picture of the squirrel when it is 7.50 m away from the tree in the direction 15.0◦ E of S. What is the magnitude and direction of the average velocity of the squirrel between the two pictures?

Answer 1

Here for average velocity we have to take vector addition into consideration,
So for the first picture we have given,
x1 = [12.0cos30.0º i + 12.0sin30.0º j] m = [10.4 i + 6.00 j] m
Now after 20 sec we have second picture whose displacement can be written as,
x2 = [7.50cos-75.0º i + 7.50sin-75.0º j] m = [1.94 i - 7.24 j] m

Here the angle we have taken clockwise form the east or the x axis and due to clockwise direction we have to take it negative

So angle will be theta = 15-90° = -75 ° , which i have used above for second picture.

So that total displacement is given as, S  = x2 - x1 = [-8.46 i - 13.2 j] m

Now the average velocity is given as,

Vav = total displacement/ total time taken = S/delta t = (-8.46i-13.2j)/20-0

= - 0.423 i - 0.66 j

Hence the magnitude of average velocity will be given as,

|Vav| = (0.423^2 + 0.66^2) ^0.5 = 0.7839 m/s ,

Which is the required magnitude of the average velocity

And direction will be . = arctan(-0.66/-0.423) = 57.3438 ° + 180º (to get into Quator III) = 237º
which is 57.3º S of West and 32.7º W of South. Which is the required direction.