Question

1. a bus moving at $v_0 = 20.0$ m/s slows down at $-2.00$ m/s² until it stops. what stopping distance does this require? find d

Explanation:

Step1: List known variables

Initial velocity $v_0 = 20.0\ \text{m/s}$, final velocity $v = 0\ \text{m/s}$, acceleration $a = -2.00\ \text{m/s}^2$

Step2: Select kinematic equation

Use $v^2 = v_0^2 + 2ax$ to solve for distance $x$

Step3: Rearrange for stopping distance

$x = \frac{v^2 - v_0^2}{2a}$

Step4: Substitute values into formula

$x = \frac{0^2 - (20.0)^2}{2\times(-2.00)} = \frac{-400}{-4.00}$

Answer:

$100\ \text{m}$