Question

You decide to travel by car for your holiday visits this year. You leave early in the morning to avoid congestion on the roads. This enables you to drive at a comfortable speed of v1=65.9 mphfor t1=3.22 hours. However, after this time, you unexpectedly come to a stop for tstop=42.1 min. Traffic starts moving again and you finish your travel at v2=57.6 mph for an additional t2=0.95 hours. How far did you travel on this trip? What was your average speed? There are 1609 meters in one mile.

Answer 1

Answer:

$d_t=266.918\ miles$ is the total distance of the trip travelled.

$s_{avg}=54.7895\ mph$ is the average speed of the journey

Explanation:

Given:

• duration of first interval, $t_1=3.22\ hr$

• speed during the first interval, $v_1=65.9\ mph$

• duration of stoppage, $t_0=42.1\ min=0.7017\ hr$

• duration of second interval, $t_2=0.95\ hr$

• speed during second interval, $v_2=57.6\ mph$

Now the total distance travelled:

$\rm distance=speed\times time$

$d_t=v_2.t_2+v_1.t_1$

$d_t=57.6\times 0.95+65.9\times 3.22$

$d_t=266.918\ miles$

The average speed:

$\rm speed_{avg}=\frac{total\ distance}{total\ time\ taken}$

$s_{avg}=\frac{d_t}{(t_0+t_1+t_2)}$

$s_{avg}=\frac{266.918}{(0.7017+3.22+0.95)}$

$s_{avg}=54.7895\ mph$