Respuesta :
Answer:
He traveled at 72km/hr during the rain
Step-by-step explanation:
Question is not well formatted. See comment
Given
[tex]d=400km[/tex] --- distance
[tex]t = 11hrs[/tex] --- total time
Let his speed from city A till the rain starts on his return trip be [tex]s_1[/tex]
Let his speed from city during the rain be [tex]s_2[/tex]
So:
[tex]s_2 = s_1 -20[/tex]
Required
Calculate [tex]s_2[/tex]
From the question, we understand that; he drives the whole 400 km and 2/5 of 400 km at [tex]s_1[/tex]
The distance covered during this period is:
[tex]d_1 = 400 + \frac{2}{5} * 400[/tex]
[tex]d_1 = 400 + 160[/tex]
[tex]d_1 = 560[/tex]
And the time during this period is:
[tex]t_1 = \frac{2}{5} * 11[/tex]
[tex]t_1 = 4.4[/tex]
So, the distance during the rain is:
[tex]d_2 = 2 * 400 - d_1[/tex]
[tex]d_2 = 2 * 400 - 560[/tex]
[tex]d_2 = 800 - 560[/tex]
[tex]d_2 = 240[/tex]
And the time during the rain is:
[tex]t_2 = 11 - t_1[/tex]
[tex]t_2 = 11 - 4.4[/tex]
[tex]t_2 = 6.6[/tex]
So, we have:
[tex]d_1 = 560[/tex] --- distance covered before the rain
[tex]d_2 = 240[/tex] --- distance covered when raining
[tex]s_2 = s_1 -20[/tex]
[tex]t_1 = 4.4[/tex] ---- time spent before the rain
[tex]t_2 = 6.6[/tex] --- time spent in the rain
Speed is calculated as:
[tex]Speed = \frac{distance}{time}[/tex]
Make distance the subject
[tex]distance = speed * time[/tex]
So:
[tex]d_1 + d_2 = s_1 * t_1 + s_2 * t_2[/tex]
Recall that:
[tex]s_2 = s_1 -20[/tex]
Make [tex]s_1[/tex] the subject
[tex]s_1 = s_2 + 20[/tex]
The expression [tex]d_1 + d_2 = s_1 * t_1 + s_2 * t_2[/tex] becomes:
[tex]560 + 240 = (s_2 + 20) * 4.4 + s_2 * 6.6[/tex]
[tex]800 = 4.4s_2 + 88+ 6.6s_2[/tex]
Collect like terms
[tex]6.6s_2 + 4.4s_2 = 880 - 88[/tex]
[tex]11s_2 = 792[/tex]
Solve for [tex]s_2[/tex]
[tex]s_2= \frac{792}{11}[/tex]
[tex]s_2=72km/h[/tex]
The distance and time 400 km and 11 hours as well as the speed
reduction gives the speed during the rainy part as 60 km/h.
Which method can be used to find the speed in the rain?
Based on the comment in the question, we have;
Given:
Distance between City A and City B = 400 km
Distance travelled at the same speed on his way back = [tex]\mathbf{\frac{2}{5}}[/tex] of the distance
Amount by by which his speed is reduced in the remaining [tex]\frac{3}{5}[/tex] of the distance because of rain = 20 km/h
The duration of the trip = 11 hours
Required:
How fast Mr. Jones was driving during the rainy part of his trip from City
B to City A?
Solution:
Let v represent the constant speed, we have;
[tex]Time = \mathbf{ \dfrac{Distance}{Speed}}[/tex]
The distance for which his speed is v = ([tex]\frac{2}{5}[/tex] + 1) × 400 km = 560 km
Total distance of the round trip = 400 km + 400 km = 800 km
The distance for which the speed = v - 20 = 800 km - 560 km = 240 km
Therefore, we have;
[tex]\mathbf{\dfrac{560}{v} + \dfrac{240}{v - 20}}= 11[/tex]
Which gives;
[tex]\mathbf{\dfrac{560 \times (v - 20) + 240 \cdot v}{v\times (v - 20) } } = 11[/tex]
11 × (v² - 20·v) = 560·v - 20 × 560 + 240·v
11·v² - 220·v - 560·v - 240·v + 11200
11·v² - 1020·v + 11,200 = 0
Using the quadratic formula, to solve the above quadratic equation we have;
[tex]v = \dfrac{1020 \pm\sqrt{(-1020)^2 - 4 \times 11 \times 11200} }{2 \times 11} = \dfrac{1020 \pm740 }{22} = \mathbf{ \dfrac{1020 \pm740 }{22}}[/tex]
v = 80 or v = [tex]12 . \overline{72}[/tex]
The possible value for the constant speed is therefore;
v = 80 km/h
The speed during the rain = v - 20, which gives;
- The speed during the rain = 80 km/h - 20 km/h = 60 km/h
Learn more about quadratic formula here;
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