a woman drove a distance of 240km on the return journey she drove 20km/h faster and took two hnurs less. what was her speed on the outward journey?
speed on first trip --- x km/h
speed on return trip -- x+2 km/h
recall that time = distance/rate
difference in their times = 2 hrs
----> 240/x - 240/(x+20) = 2
everybody times x(x+20)
240(x+20) - 240x = 2x(x+20)
240x + 4800 - 240x = 2x^2 + 40x
x^2 + 20x - 2400 = 0
(x-40)(x + 60) = 0
x = 40 or x is a negative speed -->reject
She went 40 km/h on her outward trip
and 60 km/h on her way back.
Good
To find the speed on the outward journey, we need to set up an equation using the given information.
Let's assume the speed on the outward journey is 'x' km/h. Since the distance traveled on both the outward and return journeys is the same (240 km), we can use the formula: Distance = Speed * Time.
On the outward journey:
Distance = Speed * Time
240 = x * T ------(1)
On the return journey:
Distance = (x + 20) * (T - 2) ------(2)
From equation (1), we can express Time (T) in terms of the speed (x):
T = 240 / x
Substituting this value of T into equation (2):
240 = (x + 20) * ((240 / x) - 2)
Simplifying the equation:
240x = (x + 20) * (240 - 2x)
240x = 240(x + 20) - 2x(x + 20)
240x = 240x + 4800 - 2x^2 - 40x
Simplifying further:
0 = - 2x^2 + 40x + 4800
2x^2 - 40x - 4800 = 0
x^2 - 20x - 2400 = 0
Now, we can solve the quadratic equation to find the value of x.
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
where a = 1, b = -20, c = -2400
Calculating the values using the quadratic formula:
x = (-(-20) ± √((-20)^2 - 4 * 1 * -2400)) / (2 * 1)
x = (20 ± √(400 + 9600)) / 2
x = (20 ± √10000) / 2
x = (20 ± 100) / 2
Therefore, we have two potential solutions for x:
x = (20 + 100) / 2 = 120 / 2 = 60
x = (20 - 100) / 2 = -80 / 2 = -40
Since speed can't be negative, we exclude x = -40. Hence, the speed on the outward journey is 60 km/h.