Question

A bullet hits a wall with a velocity of 20m/s and penetrates it upto a distance of 5m/s. Find the time taken by it to stop.

Answer 1

Answer:

t=0.5 seconds

Explanation:

initial velocity of bullet=20m/s

final velocity =0m/s

distance penetrated=5m

time taken =?

formula..

find acceleration first

v^2=u^2+2as

0=20^2+2a*5

0=400+10a

-400=10a

a=-40m/s^2

time=?

v=u+at

0=20-40t

40t=20

t=20/40

t=0.5 seconds

Answer 2

Small error in question " distance of 5 m or other unit but here taking 5 m "

Answer:

0.5 sec

Explanation:

Given :

Initial velocity ( u ) =  20 m /sec .

Distance ( s ) = 5 m .

Final velocity ( v ) = 0 m / sec .

From third equation of motion :

v² = u² + 2 a s

a  =  - 400 / 2 × 5

a =   - 400 / 10

a =  - 40 m / sec²

Negative sign show deacceleration :

We have to find time :

From first equation we have :

v = u + a t

0 = 20 - 40 t

40 t = 20

t = 0.5 sec .

Hence we get time 0.5 sec .