Question

WILL GIVE BRAINLIEST

Answer 1

Answer:

Step-bysteam boat (b) = speed of 36.8km/h. In 1.75 hours out same port and going same direction departed speed boat (s). In 1.8 hours later distance between the speed boat (s) and steam boat (b) was 86.9km

Scenario A) the steam boat (b) stays ahead of the speed boat (s)

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Start by first finding the total distance (d) that the steam boat (b) traveled. That would be 1.75+1.8 = 3.55 hrs

distance (d) = rate (r) × time (t)

d(s) = rt = (36.8km/h)(3.55h) = 130.64 km

Secondly, let's subtract from d the distance between b and speed boat (s):

d(s) = 130.64 - 86.9km = 43.74 km

Thirdly, since s traveled 43.74 km and left 1.8 hrs ago, let's set-up the rate of b [r(b)] equal to that distance [d(s)] divided by the time 1.8 hrs.

**Since d = rt, r = d/t

r(s) = d(s)/t = 43.74km/1.8h = 24.3km/h

Scenario B) the speed boat (s) moves ahead of the steam boat (b), this is more likely!!

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b moves before s starts for 1.75 hrs, going 36.8km/h, that distance ahead + distance it travels after the next 1.8 hrs = 130.64 km

Now s starts and travels d(s) km at r(s) km/h

If s speeds past b in 1.8 hrs, and is AHEAD by the given distance of 86.9km, then total distance of s [d(s)] = the 130.64 b traveled PLUS the 86.9 it is now ahead

d(s) = 130.64+86.9 = 217.54 km

s still only traveled that in 1.8 hrs, so the speed/rate of s is:

r(s) = d(s)/t = 217.54km/1.8h = 120.86km/h

** I think that makes more sense, that a speed boat would move MUCH faster than a steam boat, if 120.9... than over 3 times faster!!

Answer 2

Steam boat (b) = speed of 36.8km/h. In 1.75 hours out same port and going same direction departed speed boat (s). In 1.8 hours later distance between the speed boat (s) and steam boat (b) was 86.9km

Scenario A) the steam boat (b) stays ahead of the speed boat (s)
-----------------------------------
Start by first finding the total distance (d) that the steam boat (b) traveled. That would be 1.75+1.8 = 3.55 hrs
distance (d) = rate (r) × time (t)
d(s) = rt = (36.8km/h)(3.55h) = 130.64 km
Secondly, let's subtract from d the distance between b and speed boat (s):
d(s) = 130.64 - 86.9km = 43.74 km
Thirdly, since s traveled 43.74 km and left 1.8 hrs ago, let's set-up the rate of b [r(b)] equal to that distance [d(s)] divided by the time 1.8 hrs.
**Since d = rt, r = d/t
r(s) = d(s)/t = 43.74km/1.8h = 24.3km/h

Scenario B) the speed boat (s) moves ahead of the steam boat (b), this is more likely!!
-------------------------------------
b moves before s starts for 1.75 hrs, going 36.8km/h, that distance ahead + distance it travels after the next 1.8 hrs = 130.64 km
Now s starts and travels d(s) km at r(s) km/h
If s speeds past b in 1.8 hrs, and is AHEAD by the given distance of 86.9km, then total distance of s [d(s)] = the 130.64 b traveled PLUS the 86.9 it is now ahead
d(s) = 130.64+86.9 = 217.54 km
s still only traveled that in 1.8 hrs, so the speed/rate of s is:
r(s) = d(s)/t = 217.54km/1.8h = 120.86km/h

** I think that makes more sense, that a speed boat would move MUCH faster than a steam boat, if 120.9... than over 3 times faster!!