Annie is walking to school leisurely at a speed of 1.0 m/s for the first half of her trip. She then remembers that her physics teacher told her that she must average 2.5 m/s to make it to school on time. Realizing she might be late, she starts running. However, after a quick calculation, she stops running--she's going to be late anyways.
a) show why she can't make it on time
Vavg = (V1 + V2) / 2,
V1 = velocity for 1st half of trip,
V2 = velocity for 2nd half of trip,
2.5 m/s = (1 m/s + V2) / 2,
Cross multiply:
1 + V2 = 5,
V2 = 5 - 1 = 4 m/s = 9 mi/hr.
She will have to average 4 m/s or
9 mi/hr for the 2nd half of trip.
Quite a task!
Solve for V2:
To show why Annie cannot make it to school on time, we can calculate the time it would take for her to reach school based on her initial speed of 1.0 m/s for the first half of her trip, and the remaining distance she needs to cover at an average speed of 2.5 m/s.
Let's assume the total distance from Annie's starting point to school is represented by 'd'. Since she walks for the first half leisurely at 1.0 m/s, the distance covered is 'd/2'.
Now, Annie needs to calculate the remaining distance, which is also 'd/2', and the time it would take her to cover it at an average speed of 2.5 m/s.
Time taken for the second half of the trip = (remaining distance) / (average speed)
= (d/2) / (2.5 m/s)
= (1/2) * (d/2.5) seconds
= (1/2.5) * (d/2) seconds
= (2/5) * (d/2) seconds
= (1/5) * d seconds
Since Annie was already walking for the first half, the total time taken would be the sum of the time taken for the first half (d/2 m/s) and the time taken for the second half ((1/5) * d m/s).
Total time taken = (d/2) + (1/5) * d
= (5d/10) + (2d/10)
= (7d/10) seconds
So, the total time taken by Annie to reach school would be (7d/10) seconds.
Since Annie realizes that she would be late even after running, it implies that the total time taken (7d/10 seconds) is greater than the time she has available to reach school on time. Therefore, she cannot make it to school on time.