1. **State the problem:** A bus is moving at an initial velocity of $20$ m/s and comes to a stop in $4$ seconds after the brakes are applied. We need to find the distance the bus traveled before stopping. 2. **Relevant formula:** We use the equation for distance traveled under constant acceleration (or deceleration): $$ s = ut + \frac{1}{2}at^2 $$ where: - $s$ is the distance traveled, - $u$ is the initial velocity, - $a$ is the acceleration (negative for deceleration), - $t$ is the time. 3. **Find acceleration:** Since the bus stops, final velocity $v = 0$. Using the formula: $$ v = u + at $$ $$ 0 = 20 + a \times 4 $$ $$ a = \frac{0 - 20}{4} = -5 \text{ m/s}^2 $$ 4. **Calculate distance:** Substitute $u=20$, $a=-5$, and $t=4$ into the distance formula: $$ s = 20 \times 4 + \frac{1}{2} \times (-5) \times 4^2 $$ $$ s = 80 + \frac{1}{2} \times (-5) \times 16 $$ $$ s = 80 - 40 $$ $$ s = 40 \text{ meters} $$ **Final answer:** The bus traveled $40$ meters before stopping.