## Related questions with answers

Two cars leave a town at the same time and travel at constant speeds along straight roads that meet at an angle of $60^{\circ}$ in the town. If one car travels twice as fast as the other and the distance between them increases at the rate of 45 mi/h, how fast is the slower car traveling?

Solution

VerifiedIn the following figure we have considered that the slower car is moving along the x axis and the faster travels along the line $y=\sqrt{3}$

Since both cars are moving at constant speeds, we can consider that both cars were at the same time in the origin. Then their positions at time $t$ are

$\begin{aligned} \begin{array} { l } { s _ { 1 } ( t ) = v _ { 1 } t } \\ { s _ { 2 } ( t ) = 2 v _ { 1 } t } \end{array} \end{aligned}$

At time $t$, the slower car is at $P(v_1t,0)$ and the faster car at $Q(v_1 t,\sqrt{3}v_1 t)$. The distance between the objects at time t result

$\begin{aligned} d = \sqrt { \left( v _ { 1 } t - v _ { 1 } t \right) ^ { 2 } + \left( \sqrt { 3 } v _ { 1 } t - 0 \right) ^ { 2 } } = \sqrt { 3 } v _ { 1 } t \end{aligned}$

and its rate of change with respect to time is

$\begin{aligned} d ^ { \prime } ( t ) = \left( \sqrt { 3 } v _ { 1 } t \right) ^ { \prime} = \sqrt { 3 } v_ { 1 } = 45 mi/h \end{aligned}$

From this we have that

$\begin{aligned} v _ { 1 } = \frac { 45 } { \sqrt { 3 } } m i / h = 15 \sqrt { 3 }mi/h \end{aligned}$

Finally the slower car is moving with speed $v_1=15\sqrt{3}mi/h$.

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