A bullet flying at a speed of V = 400 m / s hits an earthen rampart and penetrates it to a depth of S = 0.2 m.

A bullet flying at a speed of V = 400 m / s hits an earthen rampart and penetrates it to a depth of S = 0.2 m. Determine the acceleration (a) with which it moved and the time (t)

1) It is known that a bullet flying at a speed of Vо = 400 m / s hits an earthen rampart and, having penetrated it to a depth of S = 0.2 m, stops, that is, V = 0 m / s. To determine the deceleration time of the bullet t in the earth shaft, we use the formula for the path for uniformly accelerated motion: S = (V + Vо) • t / 2; then t = 2 • S: (V + Vо). Let us substitute the values ​​of physical quantities from the condition of the problem and make calculations: t = 2 • 0.2 m: (0 m / s + 400 m / s) = 0.001 s.
2) In order to determine the acceleration with which the bullet moved in the earth shaft, we will use the formula to determine the acceleration a = Δv: t, where the change in velocity is Δv = V – Vо m / s; Δv = 0 m / s – 400 m / s = – 400 m / s and the time during which this change in speed occurs t = 0.001 s. a = – 400 m / s: 0.001s = – 400,000 m / s. The minus sign indicates an equally slow process.
Answer: t = 0.001s; a = – 400,000 m / s.