Participants in a road race take water from a refreshment station and throw their empty cups away farther down the course. If a runner has a forward speed of 6.20m/s, how far in advance of a garbage pail should he realease his water cup if the vertical distance between the lid of the garbage can and the runner's point of release is 0.50m?
pls help. When i try to solve this problem i first find the t which is time and i got 0.32seconds and then i used d=vt formula to find the d and I got 1.984m. i do not know if it is right. pls help me.
To solve this problem, you are on the right track by finding the time it takes for the runner to reach the garbage pail. However, your calculation for time seems to be incorrect. Let's go through the steps one by one:
Step 1: Find the time it takes for the runner to reach the garbage pail.
To find the time, we can use the formula: distance = speed × time.
In this case, since the distance is 0.50m and the speed is 6.20m/s, we can rearrange the formula to solve for time:
time = distance / speed
Plugging in the numbers:
time = 0.50m / 6.20m/s
time ≈ 0.081 seconds
So, it takes approximately 0.081 seconds for the runner to reach the garbage pail.
Step 2: Calculate the distance the runner should release the water cup in advance.
Now that we know the time, we can use the formula: distance = speed × time.
In this case, since the speed is 6.20m/s and we want to find the distance, we can rearrange the formula:
distance = speed × time
Plugging in the numbers:
distance = 6.20m/s × 0.081s
distance ≈ 0.503 meters
So, the runner should release the water cup approximately 0.503 meters in advance of the garbage pail.
Therefore, your calculation for the distance is almost correct. The correct distance is approximately 0.503 meters, which rounded to three decimal places is 1.984 meters.