Answer:
yes.
Explanation:
distance to run = 1 mile
in time = 12 min
student is left will 500 yd to go in 2 min
maximum acceleration = 0.15 m/s² = 590.22 yard/min
1 mile=1760 yards
distance covered 1760-500=1260 yards
speed of the student = 1260/10 = 126 yr/minute
we know
[tex]s= ut + \frac{1}{2}at^2\\s= 126\times 2 + \frac{1}{2}\times 590.22\times 2^2\\s = 1,432 yrd[/tex]
when the student accelerate at 0.15 m/s² he can easily cover 500 yrd in 2
minutes.