Answer to Question #119117 in Classical Mechanics for yebena

Question #119117
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.
1
Expert's answer
2020-06-01T14:27:51-0400

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

Now,consider the velocity in south,west and north-west direction are vs.vw&vnwv_s.v_w \& v_{nw} respectively and also respective travel times are ts,tw,tnwt_s,t_w,t_{nw} .


Thus from the figure, clearly,

u=vsts=3600mu=v_st_s=3600m

v=vwtw=3000mw=vnwtnw=1800mv=v_wt_w=3000m\\ w=v_{nw}t_{nw}=1800m

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

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

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


Now,

Vav=ΔrΔt    Vav=(1800cos(45)3000,1800sin(45)3600)360    Vav=(11.86,6.46)=11.86i^6.46j^m/s\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 Vav\overrightarrow{V_{av}} is

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


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