An archer shoots an arrow into a piece of wood. The arrow is traveling at 120 km/h when it strikes the wood. The arrow penetrates 3.8 cm into the wood before stopping. What is its average acceleration (in m/s2) into the wood?

vf^2=vi^2+2ad solve for a. vi needs to be in m/s, d in meters.

To find the average acceleration of the arrow, we need to know its initial velocity, final velocity, and the time it takes to stop.

First, let's convert the initial velocity from km/h to m/s. We know that 1 km/h is equal to 0.2778 m/s. So, the initial velocity of the arrow is:

120 km/h * 0.2778 m/s = 33.33 m/s (rounded to two decimal places)

Next, we need to find the final velocity, which is 0 m/s because the arrow stops.

Now, we need to find the time it takes for the arrow to stop. Since we only have the distance and initial velocity, we can use the formula:

(vf^2 - vi^2) = 2 * a * d

where vf is the final velocity, vi is the initial velocity, a is the average acceleration, and d is the distance.

Rearranging the formula to solve for acceleration (a), we have:

a = (vf^2 - vi^2) / (2 * d)

Plugging in the values we know:

a = (0^2 - 33.33^2) / (2 * 0.038 m)

Calculating this equation, we find:

a = -33,313.78 / 0.076 m

Finally, by dividing these numbers, we get the average acceleration:

a ≈ -437,223.42 m/s^2 (rounded to two decimal places).

Therefore, the average acceleration of the arrow into the wood is approximately -437,223.42 m/s^2.