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Jacob leaves his summer cottage and drives home. After driving for 5 hours, he is 112 km from home, and after 7 hours, he is 15
Question
Jacob leaves his summer cottage and drives home. After
driving for 5 hours, he is 112 km from home, and after 7 hours,
he is 15 km from home. Assume that the distance from home
and the number of hours driven form a linear relationship.
How long had Jacob been driving when he was 209 km from
home?
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Mathematics
5 years
2021-07-25T16:13:33+00:00
2021-07-25T16:13:33+00:00 1 Answers
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Answers ( )
Answer:
A linear relationship can be written as:
y = a*x + b
Where a is the slope and b is the y-intercept.
If this line passes through the points (a, b) and (c, d) then the slope can be written as:
a = (a – c)/(b – d)
Here y will represent the distance between Jacob and his house, and the variable x represents the time that he has ben driving.
In this case, we know that after driving for 5 hours, he is 112km from home.
Then we can write this point as (5h, 112km)
We also know that after 7 hours he is 15km from home.
Then we can write this point as (7h, 15km)
Then the slope of this function will be:
a = (15km – 112km)/(7h – 5h) = -48.5 km/h
Then the equation is:
y = -(48.5 km/h)*x + b
To find the value of b, we can replace the values of one of the points, for example in the point (7h, 15km)
This means that we need to replace x by 12h, and y by 15km, then we get:
15km = -( 48.5 km/h)*7h + b
15km + ( 48.5 km/h)*7h = b = 354.5 km
then the equation will be:
y = (-48.5 km/h)*x + 354.5 km
Now we want to answer: How long had Jacob been driving when he was 209 km from home?
Then we need to only replace y by 209km, and solve for x:
209km = (-48.5 km/h)*x + 354.5 km
209km – 354.5 km = (-48.5 km/h)*x
-145.5km = (-48.5 km/h)*x
-145.5km/( -48.5 km/h) = x = 3h
So he is 209km away from his home after driving for 3 hours.