Answer to Question #219084 in Physics for ran

Question

7. A student runs at 4.5 m/s [27 degrees S of W] for 3.0 minutes and then he turns and runs at 3.5 m\%; (35^ S of E[ for 4.1 minutes.

What was his average speed?

What was his displacement？

Answer 1

1)

$$v_{av}=\frac{3(60)(4.5)+4.1(60)(3.5)}{3(60)+4.1(60)}=3.9\frac{m}{s}$$

2)

$$d_1=3(60)(4.5)=810\ m\\d_2=4.1(60)(3.5)=861\ m\\ s_x=-861\cos{35}+810\cos{27}=16\ m\\ s_y=861\sin{35}+810\sin{27}=862\ m$$

$$s=\sqrt{862^2+16^2}=862\ m\\ \theta=\arctan{\frac{16}{862}}=1.1\degree\text{ W of S}$$