Problem:
A student driver is taking up a practical driving course to get his first driver’s
license. On the day of his practical driving, the car that he is driving with a constant
speed entered a 400-meter-long tunnel, at the same time with another car who is also
travelling with a constant speed. Due to his cautious driving first time, he is driving
slower than the other car, and it is determined that the faster car has a speed twice than
that of him. It took him 20 seconds longer to fully exit the tunnel than the other car. How
long did it take him inside the tunnel? At what constant speed is he driving the car?
What is the speed of the other car?
Solution:
Let x be the time it took the student driver to travel inside the tunnel. The formula
to find the constant speed s (in meters per second) given a covered distance d (in
meters) and the time x (in seconds) it took to cover the distance is given by:
s=d
x
Let
s1
be the speed of the student driver’s car as he traveled inside the tunnel. The
tunnel’s distance is 400 meters (d). Therefore, the speed
s1
of the student driver’s car is
given by:
s1=400
x
Let
s2
be the speed of the other car as it traveled inside the tunnel. The tunnel’s distance
is 400 meters (d). Since it took the student driver’s car 20 seconds longer to fully exit
the tunnel, then this car took (x-20) seconds inside the tunnel. Therefore, the speed
s2
of
the other car is given by:
s2=400
x−2 0
Since the faster car has a speed twice than that of the student driver’s car, then the
equation relating their speeds is given by:
s2=2s1
400
x−2 0 =2
(
400
x
)
400
x−2 0 =800
x
Solving the equation above for x to get how long did it take the student driver’s car
inside the tunnel, we get:
