Answer
(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.
Work Step by Step
(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.
