the average acceleration of the bullet through the board -55657×10⁵ m/s²
acceleration- Rate of change of velocity with respect to time.
S = 0.10 mm = 10⁻⁵ m ( distance) [∵ 1mm = 10⁻³]
u = 480 m/s (initial velocity)
v = 345 m/s (final velocity)
As we know the 3rd equation of motion.
v² = u² + 2aS
a = ? ( acceleration)
using these values in equation we get
(345)² = (480)² + 2×a× 10⁻⁵
a = (345)² - (480)² / 2× 10⁻⁵
a = -55657×10⁵ m/s²
The negative sign shows that the direction of acceleration is opposite of velocity thus bullet is slowing down.
the average acceleration of the bullet through the board -55657×10⁵ m/s²
The given question is incomplete. The complete question is.
a bullet fired straight through a board 0.10 mm thick strikes the board with a speed of 480 m/sm/s, has constant acceleration through the board, and emerges with a speed of 345 m/sm/s.
What is the average acceleration of the bullet through the board/?
to know more about the equation of motion :
brainly.com/question/13269040
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