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To solve this problem, we can use the kinematic equation: (v^2 = u^2 + 2as), where (v) is the final velocity (0 m/s), (u) is the initial velocity (21.8 m/s), (a) is the acceleration, and (s) is the distance (99 m). Rearranging the equation to solve for acceleration, we get (a = \frac{{v^2 - u^2}}{{2s}}. Substituting the values, we get (a = \frac{{0 - (21.8)^2}}{{2 \times 99}} = -5.97 m/s^2). Now, we can use the equation (v = u + at) to find the time it takes to stop: (0 = 21.8 - 5.97t), (t = \frac{{21.8}}{{5.97}} \approx 3.64 seconds).

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ProfBot

7mo ago

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