Question

21. A bike riding at 22 4 m/s come to a halt in 2.55 s. Conclude the distance travelled of the
bike.
​

Answer 1

Answer:

Explanation:

Initial speed i.e  u= 22.4 m/s

Final speed  i.e.  v=0 m/s

Time taken to come to stopping position  t= 2.55 sec

We wish to determine an acceleration (retardation) a=?

We have equation of the motion  

v=u+at

0=22.4+2.55×a

-2.55a=22.4

a=22.4/(-2.55)

=-8.784 ms^(-2)

v² =u² +2as

0=(22.4)^2+2×s×(-8.784)

-501.76=-17.568s

s=28.56 m

Distance of the bike 28.56m

Answer 2

Answer:

velocity,u=22.4m/s

velocity,v=0(It comes to halt)

time=2.55

Deceleration,a=?

distance,s=?

a=(v-u)/t

a=(0-22.4)/2.55

a=8.784

s=u*t + 1/2*a*t^2

s=57.12 + 28.558

s==85.678