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.
Answers
Answered by
2
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
Answered by
9
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 .
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