## University Physics with Modern Physics (14th Edition)

(a) $a = 249~m/s^2$ (b) The ratio of the acceleration to a freely falling body is 25.4 (c) $x = 101~m$ (d) Assuming the rate of deceleration is constant, the magazine's claim that the magnitude of acceleration is greater than 40g is not consistent with the data.
(a) $a = \frac{v}{t} = \frac{224~m/s}{0.900~s} = 249~m/s^2$ (b) $\frac{249~m/s^2}{9.80~m/s^2} = 25.4$ The ratio of the acceleration to a freely falling body is 25.4 (c) $x = \frac{1}{2}at^2 = \frac{1}{2}(249~m/s^2)(0.900~s)^2$ $x = 101~m$ (d) We can find the deceleration. $a = \frac{v-v_0}{t} = \frac{0-283~m/s}{1.40~s} = -202~m/s^2$ The magnitude of deceleration is $202~m/s^2$. We can convert this to multiples of $g$. $\frac{202~m/s^2}{9.80~m/s^2} = 20.6~g$ Assuming the rate of deceleration is constant, the magazine's claim that the magnitude of acceleration is greater than 40g is not consistent with the data.