Answer:
From the data we know that runner A and runner B are 11 km apart from the start because (6+5) km
So the runner from the east direction has distance as unknown km, rate= 9 k/h ; time= d/r=x/9 hr
So runner towards the west will be
distance = 11-x, rate= 8 k/h, time = d/r = (11-x)/8
So equating east and west time we have
x/9= (11-x)/8
8x=99-9x
17x=99
x=5.92 km
That is the distance covered by runner towards the east and he will meet the runner toward the west at
6-5.92=0.08 km west of the flagpole.
Answer:
For the aceleration we have:
Vf = Vo + a * t
Clearing "a":
a = (Vf - Vo) / t
Replacing and resolving:
a = (34 m/s - 0 m/s) / 21 s
a = 34 m/s / 21 s
a = 1,61 m/s^2
The aceleration of the vehicle is<u> 1,61 meters per second squared</u>
