Question

suzie lavtaski 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.
a. determine suzies acceleration rate. magnitude only; no +/- signs.
acceleration
$m/s/s$
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b. determine the distance suzie moved during this braking period.
distance
$m$
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1. k13q7

points: 0/1
a sports car traveling at 29.6 m/s slows to a velocity of 12.0 m/s over 13.1 s. how far does it move during this time?

Explanation:

Step1: Calculate Suzie's acceleration

Use acceleration formula: $a = \frac{v_f - v_i}{t}$
$\text{Substitute values: } v_f=3.2\ \text{m/s}, v_i=28.5\ \text{m/s}, t=3.20\ \text{s}$
$a = \frac{3.2 - 28.5}{3.20} = \frac{-25.3}{3.20} \approx 7.91\ \text{m/s}^2$ (magnitude only)

Step2: Calculate Suzie's braking distance

Use average velocity formula: $d = \frac{v_i + v_f}{2} \times t$
$\text{Substitute values: } v_f=3.2\ \text{m/s}, v_i=28.5\ \text{m/s}, t=3.20\ \text{s}$
$d = \frac{28.5 + 3.2}{2} \times 3.20 = 15.85 \times 3.20 = 50.72\ \text{m}$

Step3: Calculate sports car distance

Use average velocity formula: $d = \frac{v_i + v_f}{2} \times t$
$\text{Substitute values: } v_f=12.0\ \text{m/s}, v_i=29.6\ \text{m/s}, t=13.1\ \text{s}$
$d = \frac{29.6 + 12.0}{2} \times 13.1 = 20.8 \times 13.1 = 272.48\ \text{m}$

Answer:

a. $7.91\ \text{m/s}^2$
b. $50.72\ \text{m}$
Sports car distance: $272.48\ \text{m}$