Question

A bullet going with speed 350 m/s enters a concrete wall and penetrates a distance of 5.0 cm before coming to rest. Find the deceleration Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Solution :

Initial velocity=u=350m/s
final velcoity=v=0m/s
distance=s=5cm=5/100=0.05m
acceleration=a?
From second equation of motion :
v²-u²=2as
a=v²-u²/2s
a=o²-350x350/2x0.05
=-350x350/0.1
a= - 1225000m/s²

Hence retardation a=1225000m/s²

Answer 2

$Heya$‼

$Given \:that,$
$Initial\:velocity,\:u=350m/s$
$Final\:velocity,\:v=0m/s$

Distance covered by the bullet before coming to the rest, $d = 5 m$


$As\:we\:know,$
$Acceleration, \:a=(v^2-u^2)/2s$

$a=(0-350^2)/(2×0.05)$
$a=-122500/0.1$

$Acceleration=-1225000\:m/s^2$

$Therefore,deceleration=1225000\:m/s^2$