A bus moving at a speed of 24 m/s begins to slow at a rate of 3 m/s each second. How far does it go before stopping? (N.D.A., 2007)

### Concept & Formula This is a problem of uniform deceleration. We can use the kinematic equation of motion that relates initial velocity, final velocity, acceleration, and distance. $$v^2 - u^2 = 2as$$ Where $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration, and $s$ is distance. ### Step-by-Step Solution 1. **Identify Given Values:** * Initial velocity, $u = 24$ m/s. * Final velocity, $v = 0$ m/s (since it comes to a stop). * Acceleration, $a = -3$ m/s$^2$ (negative because it is slowing down). * Distance, $s$ is what we need to find. 2. **Apply the Kinematic Equation:** * $0^2 - (24)^2 = 2 \times (-3) \times s$ * $0 - 576 = -6s$ * $-576 = -6s$ 3. **Solve for Distance ($s$):** * $s = \frac{576}{6}$ * $s = 96$ metres. ### Exam Strategy & Shortcut Alternatively, use the concept of average speed. Time to stop = $\frac{\text{Initial Speed}}{\text{Deceleration Rate}} = \frac{24}{3} = 8$ seconds. Since deceleration is uniform, Average Speed = $\frac{Initial + Final}{2} = \frac{24 + 0}{2} = 12$ m/s. Distance = Average Speed $\times$ Time = $12 \times 8 = 96$ metres. This avoids calculating $24^2$. ### Common Pitfall Forgetting the negative sign on the deceleration rate when using the $v^2 - u^2 = 2as$ formula, which can cause confusion or lead to a negative distance result. ### Final Answer Therefore, the correct answer is **96 m**.