A student walks and jogs to college each day.She averages 5km/h walking and 9km/h jogging.Tha distance is 8km and she makes the trip in 1 hour. How far does she jog?
let her distance walking be x km
then her distance jogging is 9-x km
The total of her times walking and jogging is 1 hour
x/5 + (8-x)/9 = 1
multiply by 45
9x + 5(8-x) = 45
I will let you finish it.
To solve this problem, let's denote the distance the student walks as x km and the distance the student jogs as y km.
We are given that the total distance is 8 km, so we can write the equation:
x + y = 8 --> (Equation 1)
We are also given that the student averages 5 km/h while walking and 9 km/h while jogging. We can use the formula: speed = distance/time to find the time it takes for each segment.
For the walking segment:
speed = distance/time
5 = x/t
For the jogging segment:
speed = distance/time
9 = y/(1 - t)
Since the student spends a total of 1 hour walking and jogging, we can write:
t + (1-t) = 1
Simplifying this equation gives us:
t + 1 - t = 1
1 = 1
This means that the value of t can be any number between 0 and 1 since the equation is always true.
Now, let's solve for the distances by substituting the values of t in terms of x and y into the equations.
From the first equation, we get:
x = 8 - y
From the second equation, we get:
5 = x/t
5 = (8 - y)/t
5t = 8 - y
From the third equation, we get:
9 = y/(1 - t)
9 = y/(1 - 1 + t)
9 = y/t
Now, we have three equations:
1) x = 8 - y
2) 5t = 8 - y
3) 9 = y/t
We can solve this system of equations to find the values of x and y.
Let's substitute equation 1 into equation 2 to eliminate x:
5t = 8 - (8 - y)
5t = y
Substituting equation 3 into the above equation:
5t = 9/t
5t^2 = 9
t^2 = 9/5
t^2 = 1.8
Since t = √(1.8) can have two possible values: √1.8 and -√1.8, we need to consider both cases.
Case 1: t = √1.8
Substituting this value back into equation 3, we get:
9 = y/√1.8
y = 9√1.8
Substituting the value of y into equation 1, we get:
x = 8 - y
x = 8 - 9√1.8
Therefore, in this case, the student walks approximately 8 - 9√1.8 km and jogs approximately 9√1.8 km.
Case 2: t = -√1.8
Substituting this value back into equation 3, we get:
9 = y/-√1.8
y = -9√1.8
Substituting the value of y into equation 1, we get:
x = 8 - y
x = 8 - (-9√1.8)
x = 8 + 9√1.8
Therefore, in this case, the student walks approximately 8 + 9√1.8 km and jogs approximately -9√1.8 km.
Since distance cannot be negative, we can ignore the negative solution.
Therefore, the student jogs approximately 9√1.8 km.