Respuesta :
Answer:
Part a)
[tex]v_{max} = 7 m/s[/tex]
Part b)
[tex]d = 7 \times 3 = 21 m[/tex]
Part c)
[tex]d = 61 m[/tex]
Part d)
[tex]d = 50.5 m[/tex]
Explanation:
As we know that he start from rest and moves with uniform acceleration
so here we can say that
[tex]d = v\Delta t[/tex]
as we know here that he move with uniform speed from t = 7 to t = 11
So we will have
[tex]68 - 40 = v(11 - 7)[/tex]
[tex]28 = 4 v[/tex]
[tex]v_{max} = 7 m/s[/tex]
Part b)
If Donald runs for 3 s after reaching his maximum speed
So here distance moved by Donald
[tex]d = v t[/tex]
[tex]d = 7 \times 3 = 21 m[/tex]
Part c)
Distance of Donald from start line is given as
[tex]d = 40 + 21[/tex]
[tex]d = 61 m[/tex]
Part d)
Distance from start line after t = 8.5 s
[tex]d = 40 + vt[/tex]
[tex]d = 40 + (7)(1.5)[/tex]
[tex]d = 50.5 m[/tex]
Answer:
Maximum speed: 7m/s
Additional 3 seconds: 21m
10 seconds distance: 61 m
Distance to the starting line after 8,5= 50,5 meters
Explanation:
In order to solve this, you just have to calculate the mximum speed by withdrawing the distance covered in the first 7 seconds, from the distance covered in the 11 seconds and then divided it by the difference in time, since he ran at maximum speed from the second 7 till the end of the reace you can calulate the maximum speed easily by doing that:
68-40/11-7=28/4=7
so the maximum speed will be 7m/s
Now we multiply that by 3 seconds:
7m/s*3s= 21meters
The distance at 10 seconds after the start is given by withdrawin the first 7 seconds and adding the 40 meters to the distance of the 3 seconds at macimum speed:
40+(3s*7m/s)=40+21=61m/s
With the 8.5 we do the same but with 8.5-7=1.5 seconds
40+(1,5s*7m/s)=40+10,5=50,5m