While her kid brother is on a wooden horse at the edge of a merry-go-round, Sheila rides her bicycle parallel to its edge. The wooden horses have a tangential speed of 6 m/s. Sheila rides at 4 m/s. The radius of the merry-go-round is 8 m. At what time intervals does Sheila encounter her brother, if she rides opposite to the direction of rotation of the merry-go-round?
25.1 s
To solve this problem, we can analyze the relative speeds of Sheila and her brother, as well as the distances they cover.
First, let's determine the speed of Sheila relative to the merry-go-round. Since Sheila rides in the opposite direction of rotation, we can subtract her speed from the tangential speed of the wooden horse. Thus, the relative speed of Sheila is 6 m/s - 4 m/s = 2 m/s.
Now, we need to consider the distance-covered relationship. Since both Sheila and her brother are moving parallel to the edge of the merry-go-round, we can say that they cover the same distance in the same amount of time to meet each other.
The distance covered by Sheila's brother on the wooden horse can be calculated by multiplying the tangential speed by the time taken. Let's call this distance DB.
DB = tangential speed × time = 6 m/s × time
On the other hand, the distance covered by Sheila on her bicycle can be calculated by multiplying her relative speed by the time taken. Let's call this distance DS.
DS = relative speed × time = 2 m/s × time
Since Sheila encounters her brother when they cover the same distance, we can set up the equation:
DS = DB
2 m/s × time = 6 m/s × time
Now, we can solve for time:
2 m/s × time = 6 m/s × time
2 time = 6 time
2 = 6
Oops! It seems that the equation is contradictory and cannot be solved. This means that Sheila will never encounter her brother while riding opposite to the direction of rotation of the merry-go-round.
Therefore, we can conclude that there are no time intervals at which Sheila encounters her brother in this scenario.