Question

You bicycle 3.2 km east in 0.1 h, then 3.2 km at 15.0 degrees east of north in 0.21 h, and finally another 3.2km due east in 0.1 h to reach its destination. What is the average "velocity" for the entire trip?

Answer 1

Let the direction of displacement towards EAST be along positive direction of x-axis and that of NORTH be along positive direction of y-axis. So the unit vector along these directions be `hati and hat j` respectively.

First displacement ( `vec(d_1`) is 3.2 km east in 0.1 h

So

`vec(d_1)=3.2hatikm`

Second displacement ( `vec(d_2`) 3.2 km at 15.0 degrees east of north in 0.21 h, This means the direction of displacement makes `75^@` with `hati`

So

`vec(d_2)=(3.2cos75^@hati+3.2sin75^@hatj)km`

Third displacement ( `vec(d_3`) is 3.2 km east in 0.1 h

So

`vec(d_3)=3.2hatikm`

Hence net displacement `vec(d)`

`vec(d)=vec(d_1)+vec(d_2)+vec(d_3)`

`=[3.2hati+(3.2cos75^@hati+3.2sin75^@hatj)+3.2hati]km`

`=[6.4hati+(0.83hati+3.09j)]km`

`=[(7.23hati+3.09j)]km`

The average velocity

`v_"average"=abs(vec(d))/"Total time"=sqrt(7.23^2+3.09^2}/(0.1+0.21+0.1)(km)/(hr)`

`~~150.78"km/hr"`

And the direction of average velocity along North of East is

`theta =tan^-1(3.08/7.23)=23.14^@`