Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

30 sec

15 sec

10 sec

5 sec

Step-by-Step Solution

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.
Identify Given Values:

Acceleration $a_1 = 5 \text{ m/s}^2$
Retardation $a_2 = 10 \text{ m/s}^2$
Total Distance $S = 1.5 \text{ km} = 1500 \text{ m}$

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: $S = \frac{1}{2} \left( \frac{a_1 a_2}{a_1 + a_2} \right) T^2$
Calculation: Substitute the values into the equation: $1500 = \frac{1}{2} \left( \frac{5 \times 10}{5 + 10} \right) T^2$ $3000 = \left( \frac{50}{15} \right) T^2$ $3000 = \frac{10}{3} T^2$ $T^2 = \frac{3000 \times 3}{10} = 900$ $T = \sqrt{900} = 30 \text{ s}$
Conclusion: The minimum time required is 30 seconds .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started