Physics Page 50 # 13
In order to qualify for the finals in a racing event, a race car
must achieve an average speed of 250 km/hr on a track with a
total length of 1600 meters. If a particular car covers the
first half of the track at an average speed of 230 km/h, what
minimum average speed must it have in the second half of the
event in order to qualify?
---------------------------------------------------------------Since we have km/h and m let us convert the m into km so the
units of distance are the same.
∆X total = 1600m =
1600m 1km
•
= 1.6km
1
1000m
Since we know the total average speed needed, and the total
distance the car must travel, we can determine the total maximum
time that can be used using the distance formula.
Distance = (Rate)(Time)
∆X total = (Vmin )(t max )
(t t max ) =
∆X total
1.6km
1.6km
10 16h
8h
4h
=
=
• =
=
=
= 6.4 x10 −3 h
Vmin
250km / h 250km / h 10 2500 1250 625
Now that we know the total maximum time for the 1.6 km we can
determine how much time was used during the first 0.8km at
230km/h.
Distance = (Rate)(Time)
∆X first1/2 = (V first1 / 2 )(Time first1 / 2 )
Time first1 / 2 =
∆X first1/2
0.8km
0.8km
10
8h
2h
=
=
• =
=
≅ 3.47826 x10 −3 h
V first1 / 2
230km / h 230km / h 10 2300 575
Now that we know the total maximum time for the 1.6 km and the
time already used during the first 0.8 km, we can determine how
much time can be used for the second 0.8 km.
t max = t first1 / 2 + t sec ond 1 / 2
4h
2h
4h 575 2h 625 2300h 1250h
−
=
•
−
•
=
−
625 575 625 575 575 625 359375 359375
1050h
210h
42h
=
=
=
≅ 2.92174 x10 −3 h
359375 71875 14375
t sec ond 1 / 2 = t max − t first1 / 2 =
t sec ond 1 / 2
(Continued on the next page)
Since we now know the maximum amount of time left for the car to
cover the remaining 0.8 km, we can determine the minimum speed
that the car must be moving at to qualify for the race.
D = RT
0.8km
D
0.8km 14375 0.8km 10 14375 8km 14375 4km 14375
R= = 1 =
•
=
• •
=
•
=
•
42h
T
1
42h
1
10 42h
10
42h
5
42h
14375
2km 14375 (2km)(2875) 5750km
R=
•
=
=
≅ 273.80952km / h
5
21h
21h
21h
Vminsecond1/2 = 273.8km / h = 274km / h
NOTE: The information shown below will not work
Some students have tried to solve this problem using just average
velocities as shown below. It does not give the correct answer.
Vmin 2 qualify = Vtotalavg = 250km / h =
V first1 / 2 + Vsec ond 1 / 2
2
=
230km / h + Vsec ond 1 / 2
2
2(250km / h) = 230km / h + Vsec ond 1 / 2
500km / h = 230km / h + Vsec ond 1 / 2
Vsec ond 1 / 2 = 500km / h − 230km / h = 270km / h
Why did this method not work?
It did not work because the time for the first ½ of the course
was not the same as the time for the second ½ of the course.
For a better example imaging driving a 100 mile race. that you
had to cover in 2 hours, which means you need an average speed of
50 miles/hour. You drive the first ½ of the race at 100
miles/hour. What must be your average speed for the second half
of the race?
See what happens when we use the incorrect method shown above.
Vtotalavg = 50mi / h =
V first1 / 2 + Vsec ond 1 / 2
2
2(50mi / h) = 100mi / h + Vsec ond 1 / 2
=
100mi / h + Vsec ond 1 / 2
2
100mi / h = 100mi / h + Vsec ond 1 / 2
Vsec ond 1 / 2 = 100mi / h − 100mi / h = 0mi / h
While it is true that the average speed for 100mi/h and 0 m/h is
50 mi/h, you would never finish the race.