Two cyclists drive a distance of 15 kilometres. One of the cyclists
drives10 km h faster than the other one and arrives at their destination
20 minutes earlier. What are the speeds of the cyclists?
ok so how would i go about doing that because the vt=15 means that v is either 1,15,3,5 and likewise for t but when plugging those numbers into the first equation it does not equal 15 nor can i graph. Also where is the 1/3 from is that for the 20 mins if so why cant you just put t-20 why use a fraction. please help this is really bothering me trying to prepare fopr entrance exams after being out of high school for nearly 9 years is tough. Thank you
Let the time taken by the slower cyclist be x hrs.
Therefore, the time taken by the faster one is (x - 1/3) hrs.
The formula for speed is distance/time.
therefore,
The speed of the slower cyclist is (15/x) km/h
The speed of the faster one is [15/(x-1/3)] km/h
Since the difference between their speeds is 10 km/h,
the equation is:
[15/(x-1/3)] - (15/x) = 10
Solve for x, then calculate their speeds by substituting the value of x into their respective speeds.
If there are two values, check for both values. The value of x which gives the speed as a negative number should not be considered as an answer.
No, it doesn't meant that at all.
Units used in the question are km and km/hr. So 20 minutes has to be converted into hours for consistency.[snip]
and likewise for t but when plugging those numbers into the first equation it does not equal 15 nor can i graph. Also where is the 1/3 from is that for the 20 mins if so why cant you just put t-20 why use a fraction. please help this is really bothering me trying to prepare fopr entrance exams after being out of high school for nearly 9 years is tough. Thank you
To solve the two equations I gave you, note that from equation (2) it follows that t = 15/v. Substitute this into equation (1):
$\displaystyle \displaystyle (v + 10)\left( \frac{15}{v} - \frac{1}{3}\right) = 15$.
Expand. Simplify into a quadratic equation for v. Solve for v (only one of the solutions is acceptable). See solve v^2 + 10v - 450 = 0 - Wolfram|Alpha
and click on Show steps.
Let s = speed of the slow cyclist
then
(s+10) = speed of the faster one
:
Change 20 min to hr
:
Write a time equation: Time = dist/speed
:
slow time - fast time = hr
- =
Multiply by 3s(s+10), results:
3(15)(s-10) - 3(15)s = s(s+10)
:
45(s+10) - 45s = s(s+10)
:
45s + 450 - 45s = s^2 + 10s
:
0 = s^2 + 10s - 450
Use the quadratic formula to find s:
In this equation, x=s; a=1, b=10; c=-450
:
;
:
Positive solution
s =
s = 16.7945 ~ 16.8km/h is the slow cyclist
and
26.8 km/h is the fast cyclist
;
:
Check solution by finding the times
Slow: 15/16.8 = .89 hrs
Fast: 15/26.8 = .56 hrs
and why do you insist on sending infractions to me every time are you really that offended that i questioned your set up? it gets annoying re accepting these rules everytime.