A body travelling with a velocity of 16 metre per second square when it takes a tree trunk and comes to rest in 0.4 minutes find the time taken during the retardation
Answers
Initial velocity= u= 16 m/s
distance covered = S= 0.4 m
Final velocity=v= 0 m/s
the equation of motion;
v²=u²+2aS
0=16²+2 x a x 0.4
a=-16²/0.8
a=-320 m/s
As it carries a negative sign it represent retardation.
Time taken during the retardation is given by the equation;
v=u+at 0= 16-320 x t
t=16/320
t=1/20
t=0.05 s
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Answer:
Here is your answer...
Explanation:
The distance after which the body comes to rest is 0.4m.
The initial velocity of the body is 16m/s.
Let us say, the magnitude of retardation is a. So,
v^2=u^2+2as
0=16^2+2×a×0.4 (v=0, cause the body comes to rest.)
a = -256/0.8
a = -320m/s^2
Now, for time, we use,
v = u+at
0 = 16–320×t
-16 = -320t
16/320 = t
t = 0.05
Therefore, the time taken is 0.05s.
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