Question

A motorist drives north for $35min$ at $85km{h}^{-1}$ and then stops for $15min$. He next continues north, travelling $130km$ in $2h$.

(a) What is his total displacement?

(b) What is his average velocity?

Answer 1

Step 1: Given data

Time before stopping, $t_1=35min$

Speed before stopping, $s_1=85km{h}^{-1}$

Stopping time, $t_2=15min$

Distance covered after stopping, ${\delta }_2=130km$

Time taken after stopping, $t_3=2h$

Step 2: Find the total displacement and average velocity

(a) To find total displacement, we will use the following relation

$Speed=\frac{Dis\tan ce}{Time}$

From the above relation, distance (${\delta }_1$) is given by $Dis\tan ce=Speed\times Time$

$\Rightarrow {\delta }_1=s_1\times t_1$

$\Rightarrow {\delta }_1=85\times \frac{35}{60}$

$\Rightarrow {\delta }_1=49.58km$

and,$TotalDisplacement\left(\delta \right)={\delta }_1+{\delta }_2$

$\delta =49.58+130=179.58\approx 180km$

Step 3: Find the average velocity

(b) To find average velocity, we need total displacement and total time

$TotalTime\left(t\right)=t_1+t_2+t_3$

$t=35+15+120=170min=2h50min=2.83h$

Average velocity ($v_{av}$) is given by

$v_{av}=\frac{TotalDisplacement}{TotalTime}$

$v_{av}=\frac{\delta }{t}$

$v_{av}=\frac{180}{2.83}$

$v_{av}=63.4km{h}^{-1}$

Final Answer:

(a) The total displacement of a motorist is $\mathbf{180}\mathbf{}\mathit{k}\mathit{m}$.

(b) The average velocity of a motorist is $\mathbf{63}\mathbf{.}\mathbf{4}\mathbf{}\mathit{k}\mathit{m}{\mathit{h}}^{\mathbf{-}\mathbf{1}}$.