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27. (II) A car traveling 85km/h strikes a tree. I zuo of the car compresses and the driver comes to rest after traveling 0.80 m .What was the average acceleration of the driver during the collision? Express the answer in terms of "g's," where 1.00g=9.80m/s^2 3

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27. (II) A car traveling 85km/h strikes a tree. I zuo of the car compresses and the driver comes to rest after traveling 0.80 m .What was the average acceleration of the driver during the collision? Express the answer in terms of "g's," where 1.00g=9.80m/s^2 3

27. (II) A car traveling 85km/h strikes a tree. I zuo of the car compresses and the driver comes to rest after traveling 0.80 m .What was the average acceleration of the driver during the collision? Express the answer in terms of "g's," where 1.00g=9.80m/s^2 3

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PhilippElite · Tutor for 8 years

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First, convert the initial velocity from km/h to m/s:<br />\[ 85 \, \text{km/h} = \frac{85 \times 1000}{3600} \, \text{m/s} = 23.61 \, \text{m/s} \]<br /><br />Using the kinematic equation:<br />\[ 0 = (23.61 \, \text{m/s})^2 + 2 \cdot a \cdot 0.80 \, \text{m} \]<br />\[ 0 = 556.6321 \, \text{m}^2/\text{s}^2 + 1.6a \, \text{m} \]<br />\[ -556.6321 \, \text{m}^2/\text{s}^2 = 1.6a \, \text{m} \]<br />\[ a = \frac{-556.6321 \, \text{m}^2/\text{s}^2}{1.6 \, \text{m}} \]<br />\[ a = -347.895 \, \text{m/s}^2 \]<br /><br />Convert to "g's":<br />\[ a = \frac{-347.895 \, \text{m/s}^2}{9.80 \, \text{m/s}^2} \approx -35.5 \, g \]<br /><br />The average acceleration of the driver during the collision is approximately \(-35.5 \, g\).

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To find the average acceleration, we use the kinematic equation \(v^2 = u^2 + 2a s \). Here, \( v \) is the final velocity (0 m/s), \( u \) is the initial velocity (converted to m/s), \( a \) is the acceleration, and \( s \) is the distance (0.80 m). We then convert the acceleration to "g's".
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