Question

A bullet initially moving with a velocity of 20 m/s strikes a target and comes to reset after penetrating a distance of 0.01 m in the target calculate the retardation producer by the target.<br /><br />please help me ​

Answer 1

Given :

• Initial velocity of the bullet = 20 m/s
• Final velocity of the bullet = 0 m/s
• Distance = 0.01 m

To find :

• Retardation produced by the target

Solution :

Here, we have the value of initial velocity, final velocity and the distance and we need to calculate the retardation. So, in order to calculate it we will use the third equation of motion.

Third equation of motion :-

• v² - u² = 2as

where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• s = Distance

we have,

• v = 0 m/s
• u = 20 m/s
• s = 0.01

Substituting the given values :-

→ (0)² - (20)² = 2(a)(0.01)

→ - (20)² = 2 × a × 0.01

→ - 400 = 2 × a × 0.01

→ - 400 ÷ (2 × 0.01) = a

→ - 20,000 = a

→ Acceleration = - 20,000 m/s²

Negative acceleration is known as retardation.

So, the retardation of the bullet = - 20,000 m/s²