The graph shows the y-velocity of a 180 lb football player during a certain play

Question

The graph shows the y-velocity of a 180 lb football player during a certain play. Positive y is in the direction down the field, toward the opponents'goal. Positive x is across the field, toward the right as you face the opponents goal. There is net force in the y-direction (A) from 0-10 s ONLY from 10-20 sec ONLY from 0-10 s and from 20-30 s, but nowhere else from 0-30 sec only from 0-10 s and from 20-40 sec, and nowhere else When is there positive acceleration? from 0-10 s ONLY from 10-20 sec ONLY from 0-10 s and from 20-30 s, but nowhere else from 0-30 sec only from 0-10 s and from 20-40 sec, and nowhere else

Explanation / Answer

7) from 0-10 sec, velocity is steadily increasing(i.e it is accelerating) from 0m/s to 10m/s. Acceleration is in +xy plane i.e also in +ve y direction.

F=m*a i.e where there is a there F and both acts in same direction

8) In 0-10s interval, velocity is increasing i.e +ve acceleration. 10-20sec v=constant, a=0; 20-40sec, v is decreasing i.e -ve acceleration

9) at t =30s, velocity becomes -ve i.e he is traversing back to the starting point. At t=30 he is farthest

10) at t=30, v=0 i.e he stopped

11) a1=(10-0)/10=1m/s^2 => s1=1/2*1*10^2=50m

a2=(20-0)/10 =2m/s^2 =>s2=1/2*2*10^2=100m

i.e twice as far as in 1st case

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