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

Question

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

Accepted Answer

1. **Analyze the Motion:** To cover the distance in the minimum time, the motorcycle must accelerate at its maximum rate ($a_1$) to a peak velocity ($v_{max}$) and then immediately decelerate at its maximum rate ($a_2$) to come to rest.
2. **Identify Given Values:**
   - Acceleration $a_1 = 5 	ext{ m/s}^2$
   - Retardation $a_2 = 10 	ext{ m/s}^2$
   - Total Distance $S = 1.5 	ext{ km} = 1500 	ext{ m}$
3. **Derive Formula:**
   The relationship between total distance $S$ and total time $T$ for a particle starting and ending at rest with acceleration $a_1$ and retardation $a_2$ is:
   $$ S = \frac{1}{2} \left( \frac{a_1 a_2}{a_1 + a_2} ight) T^2 $$
4. **Calculation:**
   Substitute the values into the equation:
   $$ 1500 = \frac{1}{2} \left( \frac{5 	imes 10}{5 + 10} ight) T^2 $$
   $$ 3000 = \left( \frac{50}{15} ight) T^2 $$
   $$ 3000 = \frac{10}{3} T^2 $$
   $$ T^2 = \frac{3000 	imes 3}{10} = 900 $$
   $$ T = \sqrt{900} = 30 	ext{ s} $$
5. **Conclusion:** The minimum time required is 30 seconds .

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