A bullet travelling with a velocity of 16 m/s penetrates a tree trunk and comes to rest 0.4 m. Find the time taken during the retardation Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics
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13
Solution :
Initial velocity of bullet=16m/s
Distance=s=0.4m
Final velocity of bullet =v=0m/s
Accleration=a=?
From third equation of motion:
v²-u²=2as
a=v²-u²/2s
=0²-16²/2x0.4
=-320m/s²
Time=t?
From first equation of motion :
V=u+at
0=16-320xt
t=-16/-320=16/320=1/20=0.05sec
Hence time taken during retardation is 0.05sec
Initial velocity of bullet=16m/s
Distance=s=0.4m
Final velocity of bullet =v=0m/s
Accleration=a=?
From third equation of motion:
v²-u²=2as
a=v²-u²/2s
=0²-16²/2x0.4
=-320m/s²
Time=t?
From first equation of motion :
V=u+at
0=16-320xt
t=-16/-320=16/320=1/20=0.05sec
Hence time taken during retardation is 0.05sec
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4
HEY!!
_______________________________
▶We have initial velocity to be (u) = 16 m/s
▶Final velocity(v) = 0 m/s
▶Distance travelled against retardation =0.4 m/s
By using the equation v^2= u^2 +2as
▶0 = 16^2 +2× a × 0.4
Retardation = 256/ 0.8 = 320m/s^2
▶Time which bullet remains in the tree the= u+at
▶v = 0
▶u = 16 m/s
▶a =− 320 m/s^2
▶0 = 16 −320t
▶▶t = 16/320 = 0.05 sec
_______________________________
▶We have initial velocity to be (u) = 16 m/s
▶Final velocity(v) = 0 m/s
▶Distance travelled against retardation =0.4 m/s
By using the equation v^2= u^2 +2as
▶0 = 16^2 +2× a × 0.4
Retardation = 256/ 0.8 = 320m/s^2
▶Time which bullet remains in the tree the= u+at
▶v = 0
▶u = 16 m/s
▶a =− 320 m/s^2
▶0 = 16 −320t
▶▶t = 16/320 = 0.05 sec
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