Answer:
The bus will take 10 hours to travel from South Central High to Calgary.
Step-by-step explanation:
Let suppose that both bus and express train travels at constant speed, whose kinematic formula is now described:
[tex]v = \frac{\Delta s}{\Delta t}[/tex]
Where:
[tex]v[/tex] - Speed, measured in kilometers per hour.
[tex]\Delta s[/tex] - Travelled distance, measured in kilometers.
[tex]\Delta t[/tex] - Time, measured in hours.
The transportation time is cleared afterwards:
[tex]\Delta t = \frac{\Delta s}{v}[/tex]
The kinematic expressions for the bus and the express bus are, respectively: ([tex]\Delta s = 800\,km[/tex])
Bus
[tex]\Delta t = \frac{800\,km}{v}[/tex]
Express train
[tex]\Delta t - 2\,h = \frac{800\,km}{v + 20\,\frac{km}{h} }[/tex]
By eliminating [tex]\Delta t[/tex]:
[tex]\frac{800\,km}{v} -2\,h = \frac{800\,km}{v+20\,\frac{km}{h} }[/tex]
[tex]\frac{800\,km}{v} - \frac{800\,km}{v+20\,\frac{km}{h} } = 2\,h[/tex]
[tex]800\cdot (v+20)-800\cdot v = 2\cdot v\cdot (v+20)[/tex]
[tex]16000 = 2\cdot v^{2}+40\cdot v[/tex]
[tex]2\cdot v^{2}+40\cdot v -16000 = 0[/tex]
Which is a second-order polynomial and whose roots are:
[tex]v_{1} = 80\,\frac{km}{h}[/tex] and [tex]v_{2} = -100\,\frac{km}{h}[/tex]
Only first root is physically reasonable, as speed is a scalar, that is, a number that is represented only by magnitude. Then, the time taken by the bus to travel from Central High to Calgary is: ([tex]v = 80\,\frac{km}{h}[/tex])
[tex]\Delta t = \frac{800\,km}{80\,\frac{km}{h} }[/tex]
[tex]\Delta t = 10\,h[/tex]
The bus will take 10 hours to travel from South Central High to Calgary.
