all about physics (part2)?

Question

please give the given, formula, solution & answer. thanks :)

1. A pedestrian is running at his maximum speed of 6.0 m/s to catch a bus stopped by a traffic light. When he is 25 meters away from the bus the light changes and the bus accelerates uniformly at 1.0 m/s2. Find either; how far he has to run to catch the bus, or his frustration distance (closest approach). You may use graphs, charts, appropriate equations, tables, and/or computers.

2. A car, initially traveling with a uniform velocity, accelerates at a rate of 1.0 m/s2 for a period of 12 seconds. If the car traveled 190 m during this 12 second period, what was the velocity of the car when it started to accelerate?

3. A ball is connected to a rope and swung around in uniform circular motion. The tension in the rope is measured at 10 N and the radius of the circle is 1 m. How much work is done in one revolution around the circle?

Answer 1

1.
time to intercept=t
distance bus moves=d
pedestrian
d+25=6*t
d=6*t-25

Bus
d=1/2*1*t^2

subtract the first from the second

0=t^2-12*t+50
imaginary roots so let's look for a minima:
d/dt=2*t-12
t=6

Ped=36-25
=11m
Bus
=36/2
=18m

he gets within 7m


2.
the instantaneous velocity is
v=vi+a*t
distance is average velocity times time
d=vi*t+a*t^2/2
t=12
d=190
190=vi*12+72
vi=(190-72)/12
=9.8m/s

3. I am not sure there is work being done without acceleration


j

Answer 2

1. The pedestrian will gain on the bus until it reaches a speed of 6 m/s. This will occur after 6 seconds. At that point, the bus will have traveled a t^2 / 2 m, which equals (1) (6^2) / 2 = 18 m.

In order to catch the bus within those 6 seconds, the pedestrian has to run the 18 m the bus travels PLUS the 25 m he was behind when the bus started, or 43 m. Unfortunately, at 6 m/s he will travel only 36 m in 6 sec, so he will be 7 m short of the bus at closest approach.

2. The acceleration of 1 m/s^2 for 12 seconds adds a t^2 / 2 or (1)(12^2)/2 = 72 m to the car's progress. If the total distance traveled during this time is 190 m, the original uniform velocity has carried it 190 - 72 = 118 m in 12 seconds, indicating that the original velocity was 118/12 = 9.83333 m/s.

3. I don't understand what work is being referred to here. Mechanical work is done when a force acts through a distance. But the only force mentioned here (10 N) is not acting through a distance, so it's not clear what work is involved.