Przykład 5.
a) The plot cross the horizontal line [tex]y=2[/tex] when the time is [tex]t=5,5[/tex], so it took 5,5 s to cover the first 2 m.
b) If [tex]f(x)[/tex] denotes the distance from the starting position of the object, then its average speed over the entire 6-s period is
[tex]v_{\rm ave} = \dfrac{3\,\mathrm m - 0\,\mathrm m}{6\,\mathrm s - 0 \,\mathrm s} = \dfrac36 \dfrac{\rm m}{\rm s} = \boxed{0,5 \dfrac{\rm m}{\rm s}}[/tex]
c) In the last 3 seconds, the object covers a distance of
[tex]3\,\mathrm m - 1\,\mathrm m = \boxed{2\,\mathrm m}[/tex]
d) False. The average speed over the first 3-s period is
[tex]v_{\rm ave[0,3]} = \dfrac{1\,\mathrm m - 0\,\mathrm m}{3\,\mathrm s - 0\,\mathrm s} = \dfrac13 \dfrac{\rm m}{\rm s} \approx 0,33 \dfrac{\rm m}{\rm s}[/tex]
while over the second 3-s period, it is
[tex]v_{\rm ave[3,6]} = \dfrac{3\,\mathrm m - 1\,\mathrm m}{6\,\mathrm s - 3\,\mathrm s} = \dfrac23 \dfrac{\rm m}{\rm s} \approx 0,66\dfrac{\rm m}{\rm s} \neq 0,33\dfrac{\rm m}{\rm s}[/tex]
Przykład 2.
In total there are
8 + 24 + 28 + 16 + 4 = 80
graded assignments. Compute the percentages of students whose scores fall into the given categories:
• 0-8 : 8/80 = 1/10 = 10/100 = 10%
• 9-16 : 24/80 = 3/10 = 30/100 = 30%
• 17-24 : 28/80 = 7/20 = 35/100 = 35%
• 25-32 : 16/80 = 1/5 = 20/100 = 20%
• 33-40 : 4/80 = 1/20 = 5/100 = 5%
See the attached pie chart.
Zadanie 3.
From the plot, it appears that Mateusz
• took 6 min to reach the bus stop
• waited for 2 min
• took 5 min to return home
• took 1 min to grab his notebook
• took 5 min to return to the bus stop
• waited for 3 min
• and after the bus arrives, moves further away over the next 4 min
This means the total time Mateusz needed to (1) return home to get the notebook, (2) find the notebook, and (3) return to the bus stop is
5 min + 1 min + 5 min = 11 min
Zadanie 4.
True. Mateusz walks the distance between his house and the bus stop within the first 6 min, which is 2/5 of 1 km = 0,4 km = 400 m.
True. The bus arrives after 22 min, and its average speed is equal to Mateusz's average speed over the next 4 min. At 22 min, he is 0,4 km from home, and at 26 min, he is 4 km away from home, so the average speed is
[tex]v_{\rm ave} = \dfrac{4\,\mathrm{km} - 0,4\,\mathrm{km}}{26\,\mathrm{min} - 22\,\mathrm{min}} = \dfrac9{10} \dfrac{\rm km}{\rm min} = 0,9\dfrac{\rm km}{\rm min}[/tex]
Convert the speed to km/h.
[tex]\dfrac9{10} \dfrac{\rm km}{\rm min} \times \dfrac{60\,\rm min}{1\,\rm h} = 54 \dfrac{\rm km}{\rm h}[/tex]
