Question

When running a 100 meter race Donald reaches his maximum speed when he is 40 meters from the starting line, and 7 seconds have elapsed since the start of the race. Donald continues at this max speed for the rest of the race and is 68 meters from the starting line 11 seconds after the start of the race. What is Donald's max speed? meters per second Suppose Donald runs for an additional 3 seconds after reaching his max speed... How far will Donald travel during those 3 seconds? meters What is Donald's distance from the starting line 10 seconds after the start of the race? meters What is Donald's distance from the starting line 8.5 seconds after the start of the race? meters

Answer 1

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

Answer 2

Answer:

Part a)

$v_{max} = 7 m/s$

Part b)

$d = 7 \times 3 = 21 m$

Part c)

$d = 61 m$

Part d)

$d = 50.5 m$

Explanation:

As we know that he start from rest and moves with uniform acceleration

so here we can say that

$d = v\Delta t$

as we know here that he move with uniform speed from t = 7 to t = 11

So we will have

$68 - 40 = v(11 - 7)$

$28 = 4 v$

$v_{max} = 7 m/s$

Part b)

If Donald runs for 3 s after reaching his maximum speed

So here distance moved by Donald

$d = v t$

$d = 7 \times 3 = 21 m$

Part c)

Distance of Donald from start line is given as

$d = 40 + 21$

$d = 61 m$

Part d)

Distance from start line after t = 8.5 s

$d = 40 + vt$

$d = 40 + (7)(1.5)$

$d = 50.5 m$