Question 831866: A jet flew a distance of 4800 km. On the return flight, the speed was decreased by 200 km/h. If the difference in times was 2 hrs, what was the jets speed on the return flight
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! ---
s = d/t
d = s*t
---
outbound:
4800 = st
---
return:
4800 = (s - 200)(t + 2)
4800 = st + 2s - 200t - 400
---
st = st + 2s - 200t - 400
2s - 200t = 400
---
s = 4800/t
---
2s - 200t = 400
2(4800/t) - 200t = 400
9600/t - 200t = 400
9600 - 200tt = 400t
-200tt - 400t + 9600 = 0
---
the above quadratic equation is in standard form, with a=-200, b=-400, and c=9600
---
to solve the quadratic equation, by using the quadratic formula, copy and paste this:
-200 -400 9600
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
---
the quadratic has two real roots at:
---
x = -8
x = 6
---
the negative root doesn't fit the problem statement, so use the positive root
---
x = 6 hours
---
answer:
speed outbound:
s = 4800/6
s = 800 kph
speed on return:
s = 4800/8
s = 600 kph
---
Solve and graph linear equations:
https://sooeet.com/math/linear-equation-solver.php
---
Solve quadratic equations, quadratic formula:
https://sooeet.com/math/quadratic-formula-solver.php
---
Solve systems of linear equations up to 6-equations 6-variables:
https://sooeet.com/math/system-of-linear-equations-solver.php
|
|
|
