The engine of a motorcycle can produce a maximum acceleration $5 ext{ m/s}^2$. Its brakes can produce a maxi

Question

The engine of a motorcycle can produce a maximum acceleration $5 	ext{ m/s}^2$. Its brakes can produce a maximum retardation $10 	ext{ m/s}^2$. What is the minimum time in which it can cover a distance of $1.5 	ext{ km}$?

Accepted Answer

1.  **Analyze Motion:** To cover the distance in the minimum time starting from rest and coming to a stop (implied by the use of brakes), the motorcycle must accelerate at its maximum rate ($a_1$) to a maximum velocity ($v_{max}$) and then immediately decelerate at its maximum rate ($a_2$) to rest.
2.  **Equations:**
    *   Let $t_1$ be the acceleration time and $t_2$ be the retardation time.
    *   $v_{max} = a_1 t_1$ and $v_{max} = a_2 t_2$ (magnitude of deceleration). Thus $t_1 = \frac{v_{max}}{a_1}$ and $t_2 = \frac{v_{max}}{a_2}$.
    *   Total time $T = t_1 + t_2 = v_{max} (\frac{1}{a_1} + \frac{1}{a_2})$.
    *   Distance covered during acceleration: $S_1 = \frac{1}{2} a_1 t_1^2 = \frac{v_{max}^2}{2a_1}$ .
    *   Distance covered during retardation: $S_2 = \frac{v_{max}^2}{2a_2}$.
    *   Total distance $S = S_1 + S_2 = \frac{v_{max}^2}{2} (\frac{1}{a_1} + \frac{1}{a_2})$.
3.  **Calculation:**
    *   Given $S = 1.5 	ext{ km} = 1500 	ext{ m}$, $a_1 = 5 	ext{ m/s}^2$, $a_2 = 10 	ext{ m/s}^2$.
    *   $1500 = \frac{v_{max}^2}{2} (\frac{1}{5} + \frac{1}{10}) = \frac{v_{max}^2}{2} (\frac{3}{10})$.
    *   $1500 = v_{max}^2 (\frac{3}{20}) \implies v_{max}^2 = 10000 \implies v_{max} = 100 	ext{ m/s}$.
4.  **Find Time:**
    *   $T = 100 (\frac{1}{5} + \frac{1}{10}) = 100 (0.2 + 0.1) = 100(0.3) = 30 	ext{ s}$.

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