Question

Mr. Jones lives in City A and drives at a constant speed to City B, which is 400 km 2 of the distance at that same away. On the way back to City A, Mr. Jones drives 5 constant speed. Then, because of rain, he decreases his speed by 20 km/h. The round trip takes 11 hours. How fast did Mr. Jones drive during the rainy part of his trip from City B to City A?​

Answer 1

Answer:

He traveled at 72km/hr during the rain

Step-by-step explanation:

Question is not well formatted. See comment

Given

$d=400km$ --- distance

$t = 11hrs$ --- total time

Let his speed from city A till the rain starts on his return trip be $s_1$

Let his speed from city during the rain be $s_2$

So:

$s_2 = s_1 -20$

Required

Calculate $s_2$

From the question, we understand that; he drives the whole 400 km and 2/5 of 400 km at $s_1$

The distance covered during this period is:

$d_1 = 400 + \frac{2}{5} * 400$

$d_1 = 400 + 160$

$d_1 = 560$

And the time during this period is:

$t_1 = \frac{2}{5} * 11$

$t_1 = 4.4$

So, the distance during the rain is:

$d_2 = 2 * 400 - d_1$

$d_2 = 2 * 400 - 560$

$d_2 = 800 - 560$

$d_2 = 240$

And the time during the rain is:

$t_2 = 11 - t_1$

$t_2 = 11 - 4.4$

$t_2 = 6.6$

So, we have:

$d_1 = 560$ --- distance covered before the rain

$d_2 = 240$ --- distance covered when raining

$s_2 = s_1 -20$

$t_1 = 4.4$ ---- time spent before the rain

$t_2 = 6.6$ --- time spent in the rain

Speed is calculated as:

$Speed = \frac{distance}{time}$

Make distance the subject

$distance = speed * time$

So:

$d_1 + d_2 = s_1 * t_1 + s_2 * t_2$

Recall that:

$s_2 = s_1 -20$

Make $s_1$ the subject

$s_1 = s_2 + 20$

The expression $d_1 + d_2 = s_1 * t_1 + s_2 * t_2$ becomes:

$560 + 240 = (s_2 + 20) * 4.4 + s_2 * 6.6$

$800 = 4.4s_2 + 88+ 6.6s_2$

Collect like terms

$6.6s_2 + 4.4s_2 = 880 - 88$

$11s_2 = 792$

Solve for $s_2$

$s_2= \frac{792}{11}$

$s_2=72km/h$