Question

a bullet of mass 20 g travelling horizontally with a velocity of 180m/s and comes at rest to 0.04s calculate the distance ​

Answer 1

u =180 m/s

t =0.04 s

v=0

v = u+at

0 = 180 +a(0.04)

a=−4500m/s 2

v^2= u^2+2as

s= (v^2-u^2)2a

=  (0-32400)÷2×-4500

= 3.6m

Answer 2

Given :

• Mass, m = 20 g

• Initial velocity, u = 180 m/s

• Final velocity, v = 0 m/s

• Time, t = 0.04 seconds

To find :

• Distance, s

According to the question,

➞ v = u + at

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• t = Time

➞ Substituting the values,

➞ 0 = 180 + a × 0.04

➞ 0 - 180 = 0.04a

➞ - 180 = 0.04a

➞ - 180 ÷ 0.04 = a

➞ - 4500 = a

So,the acceleration is - 4500 m/s².

Now,

➞ s = ut + ½ at²

Where,

• s = Distance
• u = Initial velocity
• a = Acceleration
• t = Time

➞ Substituting the values,

➞ s = 180 × 0.04 + ½ × (- 4500) × 0.04 × 0.04

➞ s = 7.2 + (- 3.6)

➞ s = 7.2 - 3.6

➞ s = 3.6

So,the distance is 3.6 metres.