2 Hard Physics workbook problems! Please help! For the Holt Physics Book?

Question

1. During the Winter Olympics at calgary in 1994. Dan Jansen of the United States skated 5.00 x 10^2 m in 35.76 s. Suppose it takes Jansen 4.00 s to increase his speed from zero to his maximum speed, which is 10.0% greater than his average speed during the whole run. Calculate the magnitude of Jansen's average acceleration during the first 4.00s.
2. The winding cages in mine shafts are used to move workers in and out of the mines. These cages move much faster than any commercial elevators. In one South African mine, speeds of up to 65.0 km/h are attained. The mine has a depth of 2072 m. Suppose two cages start their downward journey at the same moment. The first cage quickly attains the maximum speed ( an unrealistic situation) then proceeds to descend uniformly at that speed all the way to the bottom. The second cage starts at rest and then increases its speed with constant acceleration of 4.00 x 10^-2 m/s/s. Which cage will reach the bottom first, and how long will it take?

Answer 1

1. The total time to skate the whole run gives us the average speed:
500=speed*35.76
speed=500/35.76

his maximum speed is
1.1*500/35.76
call this v'

since he accelerated to this speed in 4.00 seconds
v'=a*4.00
v'/4=a
plug and chug to get
3.85 m/s^2

2. This question is a bit confusing since you state that the max speed of the cage is 65.0 km/hr. If this is max and one cage attains it instantaneously, it will reach the bottom first. I don't think this is the actual question though. I think the question wants to compare a constant velocity of 65.0 km/hr to a constant acceleration of .00400 m/s^2 over a displacement of 2072 m

convert 65 km/hr to m/s
65/3.6 m/s
this cage will reach the bottom according to
2072=t1*65/3.6
t1=2072*3.6/65
or
115 seconds

The accelerating cage will reach the bottom according to
2072=.5*.004*t^2
t=sqrt(2*2072/.004)
t=1018 seconds. Much slower

j

Answer 2

Holt Physics Problem Workbook Answers