A bullet is fired through a wooden board with a thickness of 9.0 cm. The bullet

Question

A bullet is fired through a wooden board with a thickness of 9.0 cm. The bullet hits the board perpendicular to it, and with a speed of +370 m/s. The bullet then emerges on the other side of the board with a speed of +259 m/s. Assuming constant acceleration (rather, deceleration!) of the bullet while inside the wooden board, calculate the acceleration.

Calculate also the total time the bullet is in contact with the board (in sec).

Remember that since the bullet is traveling in the positive direction and it is slowing down, the acceleration is in the opposite, or negative, direction. (In this case the acceleration is negative.)

Explanation / Answer

v^2 = v0^2 + 2 a x

259^2 = 360^2 + 2*a*9.0E-2

a=-347328 m/s^2

b) v = v0 + a t

t = (v-v0)/a = (360-259)/347328= 2.91E-4 s

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