Question

1. In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 7.85 km mark at a time of 25.0 min If he accelerated for 60 s and then maintained his increased speed for the remainder of the race, calculate his acceleration over the 60 s interval. Assume his instantaneous speed at the 7.85 km mark was the same as his overall average speed up to that time.

acceleration:

2.

You are an engineer in charge of designing a new generation of elevators for a prospective upgrade to the Empire State Building. Before the state legislature votes on funding for the project, they would like you to prepare a report on the benefits of upgrading the elevators. One of the numbers that they have requested is the time it will take the elevator to go from the ground floor to the 102nd floor observatory. They are unlikely to approve the project unless the new elevators make the trip much faster than the old elevators.

If state law mandates that elevators cannot accelerate more than 3.90 m/s or travel faster than 14.8 m/s, what is the minimum time in which an elevator can travel the 373 m from the ground floor to the observatory floor?

minimum travel time:

3.

A busy chipmunk runs back and forth along a straight line of acorns that has been set out between its burrow and a nearby tree. At some instant, it moves with a velocity of −1.23 m/s. Then, 2.93 s later, it moves with a velocity of 1.97 m/s. What is the chipmunk's average acceleration during the 2.93 s time interval?

average acceleration:

Answer 1

1) the average speed is

vavg=7850m/25*60s=5.23 m/s

the distance he covers = 10-7.85=2.15 km =2150 m

the time taken by him=27 min 43.6s-25m=2 min 43.6s =163.6s

the time he has to run=163.6-60=103.6 s

now using kinematics we have

v=u+at

v=5.23+60a

again from kinematics we have

s=ut+1/2at²

s=5.23*60+1/2*a*60²

s=314+1800a...........1)

also we have

s2=v*t=(5.23+60a)*103.6

s2=542.173+6216a........2)

on adding 1 and 2 we have

s1+s2=856.173+8016a

we know that total distance s1 + s2=2150

2150-856.173=8016a

a=0.161 m/s²

so answer is 0.161 m/s² or 0.16 m/s²

2) at time t1

we have v=a*t1

the height at time t1

h1=1/2at1²

at time t2 , the height reached is

h2=1/2at1²+at1*t2

after deceleration the total height is

h3=1/2at1²+at1t2+v*t1-1/2at1²

h3=at1t2+v*t1.........1)

t1=14.8/3.90=3.79s

now using eqn 1)

h3=14.8t2+14.8*3.79487

total distnace h3=373 m

373 m=14.8t2+56.2

t2=316.8/14.8

t2=21.4 s

so the total time=2*3.79 +21.4=29 s

so answer is 29 s or 29.0 s

3) the formula for average acceleartion is

aavg=dv/dt

aavg=1.97-(-1.23) m/s/2.93-0 s=1.09 m/s²

so the answer is 1.09 m/s² or 1.1 m/s²