Acar traveling at 90 km/h strikes a tree. the front end of the car compresses and the driver comes to rest after traveling 0.72 m. what was the average acceleration of the driver during the collision? express the answer in terms of "g's," where 1.00g = 9.80 m/s2

44.29 g First, convert 90 km/h into m/s 90 km/h * 1000 m/km = 90000 m/h / 3600 s/h = 25 m/s So the car decelerated from 25 m/s to 0 m/s over a distance of 0.72 m. The formula for distance under constant acceleration is d = 1/2 A T^2 The formula for the amount of time taken to decelerate is T = V/A Substitute the formula for T into the equation for d, giving d = 1/2 A (V/A)^2 Substitute the known values of d and V into the formula 0.72 = 1/2 A (25 / A)^2 Solve for A 0.72 = 1/2 A * 625 / A^2 = 1/2 * 625 /A = 312.5/A 0.72 = 312.5/A Multiply both sides by A 0.72A = 312.5 Divide both sides by 0.72 A = 434.0278 m/s^2 Let's verify A as being correct. T will be 25 m/s divided by A giving T = 25 m/s / 434.0278 m/s^2 = 0.0576 s d = 1/2 A T^2 = 0.5 * 434.0278 * 0.0576^2 = 0.72 m We get the same stopping distance, so we verified that A = 434.0278 m/s^2 Now to convert to "g's" 434.0278 m/s^2 / 9.80 m/s^2 = 44.29 g

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