A bike accelerate uniformly from rest to a speed of 7.10m/s over a distance of 35.4 m .determine the acceleration of the bike
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Answered by
9
u=0
v=7.10m/s
S(distance)= 35.4m
a=?
v^2 - u^2=2as
7.10^2 - 0 = 2 x a x 35.4
50.41 = 70.8a
a = 70.8/50.41==> 1.4m/s^2
v=7.10m/s
S(distance)= 35.4m
a=?
v^2 - u^2=2as
7.10^2 - 0 = 2 x a x 35.4
50.41 = 70.8a
a = 70.8/50.41==> 1.4m/s^2
Answered by
3
HEYA HERE'S UR ANSWER.....
V^2-U^2=2as.
Here v=7.10m/s.
u=0.
s=35.4m.
Substitution in equation.
50.41-0=2a.35.4
50.41/35.4=2a.
1.42=2a.
1.42/2=a
0.71 m/s^2.
Hope this helps you quickly
V^2-U^2=2as.
Here v=7.10m/s.
u=0.
s=35.4m.
Substitution in equation.
50.41-0=2a.35.4
50.41/35.4=2a.
1.42=2a.
1.42/2=a
0.71 m/s^2.
Hope this helps you quickly
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