a bike riding at 22.4m/s skids to come to a halt in 2.55s.Conclude the skidding distance of the bike.
Answers
Answered by
279
solution.
initial velocity, u= 22.4 m/s
final velocity, v= 0 m/ s
Acceleration, a=?
time, t= 2.55 sec
so, a= 0- 22.4/2.55
=> a= -8.7 m/s^2
now, s= ut+ 1/2 at^2
= 22.4* 2.55+ 1/2 (-8.7)*2.55^2
= 57.12 - 28.2
= 28.92 m
______
hope it helps you☺✌✌
initial velocity, u= 22.4 m/s
final velocity, v= 0 m/ s
Acceleration, a=?
time, t= 2.55 sec
so, a= 0- 22.4/2.55
=> a= -8.7 m/s^2
now, s= ut+ 1/2 at^2
= 22.4* 2.55+ 1/2 (-8.7)*2.55^2
= 57.12 - 28.2
= 28.92 m
______
hope it helps you☺✌✌
Answered by
140
Given:-
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 28.56 m.
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