Nina, Shanti and Belle run a 1000 m race at a constant speed. When Nina crossed the finish line first, she was 200 m ahead of Shanti and 400 m ahead of Belle. When Shanthi crossed the finish line, how far ahead of Belle was she?
How did you get 5/4 oobleck?
Shanti had run 4/5 of the race when Nina finished.
So, if Nina's time was x, it took Shanti 5/4 x to finish the race
So, Bell had run 5/4 * 600 = 750 meters when Shanti finished.
That means Shanti was 250m ahead of Belle.
To Chews it 5/4 is used for Shanti because she spent x time for 800m so time for 1000m is 1000/800 x= 5/4 x
Why do you times by 600 Oobleck?
250
To Chews it. what is 1000-400
To determine how far ahead of Belle Shanti was when she crossed the finish line, we need to calculate the distance between Nina and Shanti when Shanti crossed the finish line.
Let's assume that Shanti took x seconds to complete the race. Since all three runners maintain a constant speed, we can use their relative speeds to find the distance between them at any point in time.
From the given information, we know that when Nina finished the race, she was 200 m ahead of Shanti and 400 m ahead of Belle. Therefore, when Shanti finished the race, she would be 200 m behind Nina and 400 m ahead of Belle.
We can set up two equations based on these distances:
Distance traveled by Nina = Distance traveled by Shanti + 200 m ----- (Equation 1)
Distance traveled by Nina = Distance traveled by Belle + 400 m ----- (Equation 2)
Since the three runners maintain a constant speed, we can use the formula distance = speed × time to express the distances as functions of time.
Let's assume the speeds of Nina, Shanti, and Belle are n, s, and b m/s (meters per second), respectively.
Equation 1 can be rewritten as n * x = s * x + 200 m
Equation 2 can be rewritten as n * x = b * x + 400 m
We want to find the value of (b * x) - (s * x), which represents how far ahead of Belle Shanti was when she crossed the finish line. To do this, we need to isolate (b * x) and (s * x) separately.
Rearranging Equation 1, we get n * x - s * x = 200 m
Simplifying, we find (n - s) * x = 200 m
Rearranging Equation 2, we get n * x - b * x = 400 m
Simplifying, we find (n - b) * x = 400 m
We now have a system of equations:
(n - s) * x = 200 m
(n - b) * x = 400 m
To eliminate the variable x, we can divide Equation 2 by Equation 1:
[(n - b) * x] / [(n - s) * x] = 400 m / 200 m
(n - b) / (n - s) = 2
Cross-multiplying, we get 2(n - s) = (n - b)
2n - 2s = n - b
b = n + 2s
Now we have found the relationship between the speeds of Shanti and Belle. We know that Belle finished the race after Nina and Shanti, so her time will be x + y seconds, where y is the time taken by Shanti to complete the race.
Now we can calculate how far ahead of Belle Shanti was when she crossed the finish line. We substitute b = n + 2s into Equation 2 and solve for (b * x) - (s * x):
n * x = b * x + 400 m
n * x = (n + 2s) * x + 400 m
0 = 2s * x + 400 m
400 m = 2s * x
200 m = s * x
Therefore, when Shanti crossed the finish line, she was 200 meters ahead of Belle.