A race car is one lap behind the lead race car when the lead car has 55 laps to go in a race. If the speed of the lead car is 62.3 m/s, what must be the average speed of the second car to catch the lead car just before the end of the race (i. e., right at the finish line)

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

grav potential energy

the ima (higher than 1m) will be greater than the ima with the height of 1m.