By trial and error, a frog learns that it can leap a maximum horizontal distance of 1.4 m. If, in the course of an hour, the frog spends 32% of the time resting and 68% of the time performing identical jumps of that maximum length, in a straight line, what is the distance traveled by the frog?
Yes, it's a super frog.
The time of jumping is
0.68•3600 = 2448 sec.
The distance covered by this small “projectile” (the frog) during 1 jump is
L =v²•sin2α/g.
We have Lmax => sin2α =1 and α = 45º.
Lmax= v² /g.
v =sqrt(Lmax•g) = sqrt(1.4•9.8) =3.7 m/s.
The time of 1 jump is
t = 2•v•sin α/g = 2•3.7•sin45º/9.8 =0.53 sec.
During the hour the frog made N jumps:
N = 2448/0.53=4619.
The total distance is
1.4•4619 =6467 m.
To find the distance traveled by the frog, we need to calculate the total distance covered during the jumps.
Let's assume the total time in an hour is 60 minutes. The frog spends 32% of this time resting, which is equal to 0.32 * 60 = 19.2 minutes.
The remaining time, which the frog spends jumping, is 60 - 19.2 = 40.8 minutes.
Now, we need to determine the distance covered in 40.8 minutes. We know that the frog's maximum horizontal distance for each jump is 1.4 meters.
To calculate the distance covered, we multiply the time spent jumping by the distance per jump: 40.8 minutes * 1.4 meters/minute.
However, since the time is given in minutes and the distance per jump is given in meters/minute, we get an incorrect unit of meters^2/minute.
To correct this, we need to convert the time from minutes to hours. There are 60 minutes in an hour, so 40.8 minutes is equal to 40.8/60 = 0.68 hours.
Now, we can calculate the distance traveled by the frog as follows:
Distance = Time * Distance per jump
Distance = 0.68 hours * 1.4 meters/minute
To find the distance in meters, we need to convert 0.68 hours to minutes:
0.68 hours * 60 minutes/hour = 40.8 minutes
Distance = 40.8 minutes * 1.4 meters/minute = 57.12 meters
Therefore, the frog travels a total distance of 57.12 meters.