QuestionJuly 19, 2025

A car begins braking at a speed of 23.4m/s The car stops after 5.57s. What is the car's acceleration during this time? How far does the car travel while braking? square

A car begins braking at a speed of 23.4m/s The car stops after 5.57s. What is the car's acceleration during this time? How far does the car travel while braking? square
A car begins braking at a speed of 23.4m/s The car stops after 5.57s. What is the car's acceleration during this time? How far does the car travel while braking? square

Solution
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Answer

Acceleration: -4.20 \, m/s^2, Distance: 65.1 \, m Explanation 1. Calculate the acceleration Use the formula for acceleration: **a = \frac{v_f - v_i}{t}**. Here, v_f = 0 \, m/s, v_i = 23.4 \, m/s, and t = 5.57 \, s. Thus, a = \frac{0 - 23.4}{5.57}. 2. Calculate the distance traveled Use the formula for distance: **d = v_i \cdot t + \frac{1}{2} a \cdot t^2**. Substitute v_i = 23.4 \, m/s, t = 5.57 \, s, and a from Step 1.

Explanation

1. Calculate the acceleration<br /> Use the formula for acceleration: **$a = \frac{v_f - v_i}{t}$**. Here, $v_f = 0 \, m/s$, $v_i = 23.4 \, m/s$, and $t = 5.57 \, s$. Thus, $a = \frac{0 - 23.4}{5.57}$.<br /><br />2. Calculate the distance traveled<br /> Use the formula for distance: **$d = v_i \cdot t + \frac{1}{2} a \cdot t^2$**. Substitute $v_i = 23.4 \, m/s$, $t = 5.57 \, s$, and $a$ from Step 1.
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