Solve 1st question I'M not getting the appropriate answer.URGENT................

1. A motor bike, initially at rest, picks up a velocity of 72 kmh^–1 over a distance of 40 m. Calculate 
(i) acceleration (ii) time in which it picks up above velocity. 

​

Velocity v of the bike after travelling 40 m = 72 km/ h= (72000 m)/ 3600 s = 20 m/s.

Using the relation : v² - u² = 2 a s

Here v= 20 m/s, u = 0 m/s (as the bike was at rest), s = 40 m and a = acceleration of the bike. Substituting we get,

20² - 0² = 2 a × 40; ==> 400 = 80 a; ==> a = 5 m/s²

We can obtain time from v = u + a t.

Substituting we get

20 m/s = 0 m/s +( 5m/s²) × t ;==> t = (20 m/s)/ (5m/s²)= 4s

The acceleration of the bike = 5m/s² and it takes 4 s to achieve the velocity of 20m/s ie 72 km/h.

I had earlier solved the question wrongly using v = 2 m/s instead of the correct figure of 20 m/s.