Question

1. k13q3

captain john stapp is often referred to as the \fastest man on earth.\ in the late 1940s and early 1950s, stapp ran the u.s. air forces aero med lab, pioneering research into the accelerations which humans could tolerate and the types of physiological effects which would result. manning the rocket sled on the famed gee whiz track, stapp tested acceleration and deceleration rates in both the forward-sitting and backward-sitting positions. he would accelerate to aircraft speeds along the 1200-foot track and abruptly decelerate under the influence of a hydraulic braking system. on one of his most intense runs, his sled decelerated from 269 m/s (603 mi/hr) to a stop at -200 m/s/s. determine the ...
a. ... stopping distance.
distance
m
info attempts: 3/
submit
b. ... stopping time.
time
s
info attempts: 0/∞
submit
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1. k13q5

has reached the end of the ski slope and abruptly decelerates from 28.5 m/s to 3.2 m/s in 3.20 seconds.

Explanation:

Step1: Convert initial velocity (no, wait, first find stopping time first? Wait no, first let's list known values: initial velocity $v_0 = 269$ m/s, final velocity $v = 0$ m/s, acceleration $a = -200$ m/s². First find stopping time (part b), then use that for distance (part a).

Step1: Find stopping time

Use $v = v_0 + at$. Solve for $t$:
$t = \frac{v - v_0}{a}$
$t = \frac{0 - 269}{-200}$

Step2: Calculate stopping time

$t = \frac{-269}{-200} = 1.345$ s

Step3: Find stopping distance

Use $d = v_0 t + \frac{1}{2} a t^2$
Substitute $v_0=269$, $t=1.345$, $a=-200$:
$d = 269 \times 1.345 + \frac{1}{2} \times (-200) \times (1.345)^2$

Step4: Compute each term

First term: $269 \times 1.345 = 361.805$
Second term: $\frac{1}{2} \times (-200) \times 1.809025 = -100 \times 1.809025 = -180.9025$
Sum: $361.805 - 180.9025 = 180.9025$

Answer:

a. Stopping distance: $181$ m (rounded to 3 significant figures)
b. Stopping time: $1.35$ s (rounded to 3 significant figures)