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A bus goes from town A to B in an exact time. If the bus goes at the rate of 50km/h, then it will arrive B 42min later than it had to and if it increases its speed 5.5/9 m/sec., it will arrive B 30min earlier. Find:
A) The distance between the two towns;
B) The exact time that it takes to arrive town B
C) The speed of the bus(by schedule) for the exact time.

Sagot :

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Let the speed of bus for the exact time = x km/h
the distance between the cities = y km
then the exact time would be, t hours = (y/x) hours

 If the bus goes at the rate of 50km/h, then it will arrive B 42min later,
speed = 50 km/h
42 minutes = 42/60 hours = 7/10 hours = 0.7 hours
time taken = t+0.7
distance = speed×time
⇒ y = 50×(t+0.7)
⇒ y = 50t + 35     ---------------------(1)

it increases its speed 5.5/9 m/sec, it will arrive B 30min earlier.
5.5/9 m/s = (5.5/9)×(18/5) km/h = 2.2 km/h
30 minutes = 30/60 = 0.5 hour
speed = (x+2.2) km/h
time = (t - 0.5) hours

distance = speed×time
⇒ y = (x+2.2)×(t-0.5)
⇒ y = ((y/t) +2.2)×(t-0.5)               (t = y/x)
⇒ y = y - 0.5 (y/t) + 2.2t - 1.1
⇒ 0.5 (y/t) - 2.2t + 1.1 = 0           (subtracting y from both sides)
⇒ (y/t) - 4.4t - 2.2 = 0                (dividing both sides by 0.5)
⇒ y - 4.4t² - 2.2t = 0                  (multiplying both sides by t)
⇒ 50t + 35 - 4.4t² - 2.2 t = 0      (from equation 1)
⇒ -4.4t² + 35 + 47.8t = 0
⇒ 4.4t² - 47.8t - 35 = 0

solving the quadratic equation, we get t = 11.55 hours
y = 50t + 35 = 612.5 km
x = 612.5/11.55 = 53 km/h

A) 612.5 km
B) 11.55 hours
C) 53 km/h
The Logic Defined:

1 Minute=t, (a unit of time)

Time (By schedule)=nt, (n>0), nt=number of minutes

Metre(s)=m

Speed=s (in metres per minute), s=[distance in metres]/[time in minutes]

Distance=d (in metres), d=[speed in metres per minute]*[time in minutes]

---------------------------------------------

Statement (1):

"If the bus goes at the rate of 50km/h, then it will arrive B 42min later than it had to."

Conclusion 1:

[tex]\frac { 50km }{ h } =\frac { 50,000m }{ 60t } =\frac { 2,500m }{ 3t } \\ \\ \therefore \quad \frac { 2,500m }{ 3t } =nt+42t\\ \\ \frac { 2,500m }{ 3t } =t\left( n+42 \right)[/tex]

[tex]\\ \\ 2,500m=3{ t }^{ 2 }\left( n+42 \right) \\ \\ m=\frac { 3{ t }^{ 2 }\left( n+42 \right) }{ 2,500 } [/tex]

Statement (2):

"if it increases its speed 5.5/9 m/sec., it will arrive B 30min earlier."

Conclusion 2:

[tex]\frac { 5.5m }{ 9\quad seconds } =\frac { 5.5m }{ \frac { 9 }{ 60 } t } =\frac { 110m }{ 3t }[/tex]

[tex]\\ \\ \therefore \quad \frac { 110m }{ 3t } =nt-30t\\ \\ \frac { 110m }{ 3t } =t\left( n-30 \right) \\ \\ 110m=3{ t }^{ 2 }\left( n-30 \right)[/tex]

[tex]\\ \\ m=\frac { 3{ t }^{ 2 }\left( n-30 \right) }{ 110 } [/tex]

Conclusion 3, because of conclusion 1 and 2:

[tex]\frac { 3{ t }^{ 2 }\left( n-30 \right) }{ 110 } =\frac { 3{ t }^{ 2 }\left( n+42 \right) }{ 2,500 } \\ \\ 7,500{ t }^{ 2 }\left( n-30 \right) =330{ t }^{ 2 }\left( n+42 \right) \\ \\ 7,500\left( n-30 \right) =330\left( n+42 \right)[/tex]

[tex]\\ \\ 7,500n-225,000=330n+13,860\\ \\ 7,500n-330n=13,860+225,000\\ \\ 7,170n=238,860\\ \\ n=\frac { 238,860 }{ 7,170 } \\ \\ \therefore \quad n=\frac { 7962 }{ 239 } [/tex]

Therefore,

[tex]Time\quad by\quad schedule=\frac { 7962 }{ 239 } t\\ \\ Approx:\quad 33.3\quad mins[/tex]

Now we want to find the distance between the two towns, so we say that:

[tex]d=\frac { 2,500m }{ 3t } \cdot \left( \frac { 7962 }{ 239 } t+42t \right) \\ \\ =\frac { 2,500m }{ 3t } \cdot \frac { 18,000 }{ 239 } t[/tex]

[tex]\\ \\ =\frac { 45,000,000 }{ 717 } m\\ \\ Approx:\quad 62,761.5\quad metres\\ \\ In\quad km\quad (approx):\quad 62.761\quad km[/tex]

So now you want to know how fast the bus has to travel to get to its destination on time...

Use the formula: s=d/t

Therefore:

[tex]s=\frac { \frac { 45,000,000 }{ 717 } m }{ \frac { 7962 }{ 239 } t } \\ \\ Approx:\quad 1,883.9\quad metres\quad per\quad minute[/tex]
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