Question

A cab driver heads south with a steady speed of
v1 = 22.0 m/s
for
t1 = 3.00 min,
then makes a right turn and travels at
v2 = 25.0 m/s
for
t2 = 2.60 min,
and then drives northwest at
v3 = 30.0 m/s
for
t3 = 1.00 min.
For this 6.60-min trip, calculate the following. Assume +x is in the eastward direction.total vector displacement (Enter the magnitude in m and the direction in degrees south of west.) andaverage velocity (Enter the magnitude in m/s and the direction in degrees south of west.)

Answer 1

$$\bold{d}=(0,-22)(3\cdot 60)+(-25,0)(2.6\cdot 60)+(30\cos{45\degree},-30\sin{45\degree})(60)$$

$$\bold{d}=(-2627,-5233)\ m$$

$$d=5860\ m$$

$$\theta= 63\degree$$

$$v=\frac{5860}{(3+2.6+1)60}=14.8\frac{m}{s}$$

$$\theta= 63\degree$$