You normally drive on the freeway between San Diego and Los Angeles at an average speed of 105 km/h (65 mi/h), and the trip takes 1 h and 50 min. On a Friday afternoon, however, heavy traffic slows you down and you drive the same distance at an average speed of only 70 km/h (43 mi/h). How much longer does the trip take?

• Your average speed on the freeway is,$v_{\text{1}}=\text{105km/h}$
• Time taken in the trip on the freeway is,$t_{\text{1}}=\text{1} \text{h} \text{50} \text{min}$
• Your average speed after the heavy traffic is,$v_2=\text{70} \text{km/h}$

The expression for the average velocity can be given as,

$v=\frac{d}{t}$

Where,v= average velocity, d= displacement, and t= travel time

So, for the freeway, your average speed will be,

$v_1=\frac{d}{t_1}$

Substituting the values in the above expression,

$\begin{array}{rcl}d& =& \left(105 km/h\right)\times \left(1.83 \text{h}\right)\\ & =& 192.5km\end{array}$

Therefore, the total covered distance is, $d=192.5 \text{km}$

For the traffic way

$v_2=\frac{d}{t_2}$

Substituting the values in the above expression,

$\begin{array}{rcl}{t}_2& =& \frac{192.5km}{70km/h}\\ & =& 2.75 \text{h}\times \left(\frac{60min}{1h}\right)\\ & =& 2 \text{h}45 \text{min}\end{array}$

Thus, the travel time during the heavy traffic is, $t_2=2 \text{h}45 \text{min}$

Therefore, the difference in the travel time during the heavy traffic and on a freeway can be determined by,

$\Delta t=t_2-t_1$

Substituting the values in expression will give,

$\begin{array}{rcl}\Delta t& =& 2\text{hr}\text{45}\text{min}-1\text{hr}\text{50}\text{min}\\ & =& 55\text{min}\\ & & \end{array}$

Thus, you will take 55 min longer in the heavy traffic as compared to a freeway.