Mdabhi1 wrote:
Bunuel wrote:
OfficialSolution:
To arrive at its destination on time, a bus should have maintained a speed of\(v\) kilometers per hour throughout the journey. However, after traveling the first third of the distance at\(v\) kilometers per hour, the bus increased its speed and covered the rest of the distance at\(1.2v\) kilometers per hour. As a result, the bus arrived at its destination\(x\) minutes earlier than planned. What was the actual duration of thetrip?
The bus covered\(\frac{1}{3}\) of the distance at\(v\) kilometers per hour and the remaining\(\frac{2}{3}\) of the distance at\(1.2v\) kilometers per hour.
Let the actual duration of the trip be\(t\) hours and the total distance be\(d\) kilometers. Then wehave:
\( t =\frac{(\frac{d}{3})}{v}+\frac{(\frac{d2}{3})}{1.2v}\) , which simplifies to\(t=\frac{d}{v}*(\frac{1}{3}+\frac{2}{3.6})\) , and finally to\(t=\frac{d}{v}*\frac{8}{9}\)
We also know that if the speed throughout the journey had been\(v\) kilometers per hour, the bus would have needed\(\frac{x}{60}\) hours more time to
...
To arrive at its destination on time, a bus should have maintained a speed of\(v\) kilometers per hour throughout the journey. However, after traveling the first third of the distance at\(v\) kilometers per hour, the bus increased its speed and covered the rest of the distance at\(1.2v\) kilometers per hour. As a result, the bus arrived at its destination\(x\) minutes earlier than planned. What was the actual duration of thetrip?
The bus covered\(\frac{1}{3}\) of the distance at\(v\) kilometers per hour and the remaining\(\frac{2}{3}\) of the distance at\(1.2v\) kilometers per hour.
Let the actual duration of the trip be\(t\) hours and the total distance be\(d\) kilometers. Then wehave:
\( t =\frac{(\frac{d}{3})}{v}+\frac{(\frac{d2}{3})}{1.2v}\) , which simplifies to\(t=\frac{d}{v}*(\frac{1}{3}+\frac{2}{3.6})\) , and finally to\(t=\frac{d}{v}*\frac{8}{9}\)
We also know that if the speed throughout the journey had been\(v\) kilometers per hour, the bus would have needed\(\frac{x}{60}\) hours more time to
...
Statistics : Posted by Bunuel • on Sep 16, 2014 7:41 am • Replies 8 • Views 37469
.jpg)
