Answer to Question #119117 in Classical Mechanics for yebena

Question

A motorist drives south at 20 m/s for 3 min, then turns west and travels at 25 m/s for 2 min, and finally travels northwest at 30 m/s for 1 min. For this 6 min trip find the average velocity.

Answer 1

Let the motorist starts his journey from the origin $A=(0,0)$ and reach to points $B,C,D$ as given in the problem.

Now,consider the velocity in south,west and north-west direction are $v_s.v_w \& v_{nw}$ respectively and also respective travel times are $t_s,t_w,t_{nw}$ .

Thus from the figure, clearly,

$$u=v_st_s=3600m$$

$$v=v_wt_w=3000m\\ w=v_{nw}t_{nw}=1800m$$

Thus from the figure we can clearly see the final displacement vector is

$$\overrightarrow{r}=(-1800\cos(45^{\circ})-3000,-1800\sin(45^{\circ})-3600)$$

and initial displacement vector is $(0,0)$ .

Now,

$$\overrightarrow{V_{av}}=\frac{\Delta \overrightarrow{r}}{\Delta t}\\ \implies \overrightarrow{V_{av}}= \frac{(-1800\cos(45^{\circ})-3000,1800\sin(45^{\circ})-3600)}{360}\\ \implies \overrightarrow{V_{av}}=(-11.86,-6.46)=-11.86\hat{i}-6.46\hat{j} \:m/s$$

Hence, magnitude of $\overrightarrow{V_{av}}$ is

$$V_{av}=\sqrt{(-11.86)^2+(-6.46)^2}=13.50m/s$$