Answer:
The second car must go with a speed of 63.43 m/sec
Explanation:
Speed V of lead car = 62.3 m/sec
Distance S = 55 laps = 55 ×400 meters=22000 m
We know
S = V × t
So,
t= S/V
We put values of S and V here, we get
t=22000/62.3
t= 353.1 sec
So in 353.1 sec the second car which is one lap behind - must go a distance of 55+1=56 laps or 56×400 m = 22400 meters to catch the lead car before it finishes.
i-e for second car
Distance S= 22400m
Time t = 353.1 sec
V= ?
using again
S=Vt
we get
V= S/t
V= 22400/353.1= 63.43 m/sec
Answer:
If the wind is offshore (blowing away from the dock), one should carefully approach the dock at a 20 to 30 degree angle. A bow line is then passed ashore and secured. In boats having an outboard, or inboard/outboard engine, the engine is turned towards the dock and put in reverse. This invariably will bring the stern into the dock.
Answer:
254
Explanation:
use the formula "final Velocity- initial velocity / time = acceleration"
so "X - 14 /4 = 60"
60 x 4 = X - 14
240 +14 = X
X = 254
