Question

A bullet hits a wall with a velocity of 20 m/sec and panetrates it up to a
distance of 5 cm. Find the decceleration of the bullet in the wall.
pls guys I don't understand it pls explain​

Answer 1

${\bold{\pink{\underline{\green{G}\purple{iv}\orange{en}\red{:-}}}}}$

• $\bf\:Initial\:velocity\:(u)=20\:m/s$

• $\bf\:Distance\:(s)=5\:cm$

• $\bf\:Distance\:(s)=\dfrac{5}{100}\:m$

• $\bf\:Distance\:(s)=0.05\:m$

• $\bf\:Final\:velocity\:(v)=0\:m/s$

${\bold{\blue{\underline{\red{To}\:\pink{Fin}\green{d}\purple{:-}}}}}$

• $\bf\:Deceleration\:of\:the\:bullet=?$

${\bold{\pink{\underline{\red{So}\purple{lut}\green{ion}\orange{:-}}}}}$

Now,

$\sf\boxed\star\purple{\underline{\underline{{Using\:third\:equation\:of\:motion:-}}}}$

$\bf\:{v}^{2}={u}^{2}+2as$

$\bf\implies\:{0}^{2}={20}^{2}+2\times\:a\times\:0.05$

$\bf\implies\:a=-4000\:m/{s}^{2}$

hence,

Deceleration of the bullet on the wall will be -4000 m/s².

➥Other equation of motion:-

☛ $\sf\red{{v=u+at}}$

☛ $\sf\green{{s=ut+\dfrac{1}{2}a{t}^{2}}}$

Answer 2

$\huge \boxed{ \fcolorbox{orange}{red}{Answer}}$

u= 20 m/s

v= 0

s = 5 cm = 0.05 m

from 3rd equation,

v² -u² = 2as

⇒0 - (20)² = 2×0.05×a

⇒0.1 a = - 400

⇒a = - 4000 m/s²

deceleration will be - 4000 m/s²