Question

12. A baseball player just hit the ball a long way and starting sprinting around the bases. He/she ran 90ft to first base and it took him/her 6.4 seconds to get there. As this person crossed 1st base their instantaneous velocity was 15.4 ft/sec. They rounded the base and began running another 90ft toward second base. It took them 5.1 seconds to get from 1st base to 2nd base. Still not thrown out, they proceeded to sprint another 90ft toward 3rd base. They made it from 2nd base to 3rd base in 4.9 seconds. As they crossed 3rd base their instantaneous velocity was 17.4 ft/sec. Finally realized that they were going to make it home, they sprinted the final 90ft and it took them 5.0 seconds to go from 3rd base to home plate. Their instantaneous velocity as they crossed home plate was 11.5 ft/sec. f. Average acceleration from home base to 1st base. g. Average acceleration from 3rd base to home base.

Answer 1

from home to first base

s=90 ft = 27.432 m; t= 6.4 s; v=15.4 ft/s = 4.7 m/s

using two of the equations of motion

s= ut + 0.5at²

27.432 = u*6.4 + 0.5*a*6.4²

and

v = u +at

4.7 = u + a*6.4

putting the value of 'u' from second equation to first

27.432 = (4.7 - a*6.4)*6.4 + 0.5*a*6.4²

now solving

a = 0.129 m/s² = 0.423 ft/s²

from third to home base

s=90 ft = 27.432 m; t= 5 s; v=11.5 ft/s = 3.5 m/s; u= 17.4 ft/s = 5.3 m/s

using,

v = u+at

3.5 = 5.3 + a*5

a= -0.36 m/s²

= -1.18 ft/s²