Question

Two friends, barbara and neil, are out rollerblading. with respect to the ground, barbara is skating due south at a speed of 5.9 m/s. neil is in front of her. with respect to the ground, neil is skating due west at a speed of 1.4 m/s. find neil's velocity (a) magnitude and (b) direction relative to due west, as seen by barbara.

Answer 1

As seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west. Let's assume that both Barbara and Neil start out at coordinate (0,0) and skate for exactly 1 second. Where do they end up? Barbara is going due south at 5.9 m/s, so she's at (0,-5.9) Neil is going due west at 1.4 m/s, so he's at (-1.4,0) Now to see Neil's relative motion to Barbara, compute a translation that will place Barbara back at (0,0) and apply that same translation to Neil. Adding (0,5.9) to their coordinates will do this. So the translated coordinates for Neil is now (-1.4, 5.9) and Barbara is at (0,0). The magnitude of Neil's velocity as seen by Barbara is sqrt((-1.4)^2 + 5.9^2) = sqrt(1.96 + 34.81) = sqrt(36.77) = 6.1 m/s The angle of his vector relative to due west will be atan(5.9/1.4) = atan(4.214285714) = 76.7 degrees So as seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west.

Answer 2

It is given that,

Speed of Barbara w.r.t ground, $v_b=-5.9\ m/s$ ( in south)

Speed of neil w.r.t ground, $v_n=-1.4\ m/s$ ( in west)

So, Neil's velocity as seen by Barbara is :

$v^2=\sqrt{(1.4\ m/s)^2+(5.9\ m/s)^2}$

$v=6.06\ m/s$

Direction relative to due wet is :

$tan\theta=\dfrac{5.9\ m}{1.4\ m}$

$\theta=tan^{-1} (4.21)$

$\theta=76.6\ ^0$

So, with respect to barbara, Neil is travelling with velocity of 6.06 m/s making an angle of $76.6^0$ due west.

Hence, this is the required solution.