Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The engine of a motorcycle can produce a maximum acceleration $5 \text{ m/s}^2$ . Its brakes can produce a maximum retardation $10 \text{ m/s}^2$ . 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 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.
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 $S = S_1 + S_2 = \frac{v_{max}^2}{2} (\frac{1}{a_1} + \frac{1}{a_2})$ .
Calculation: Given $S = 1.5 \text{ km} = 1500 \text{ m}$ , $a_1 = 5 \text{ m/s}^2$ , $a_2 = 10 \text{ m/s}^2$ . $1500 = \frac{v_{max}^2}{2} (\frac{1}{5} + \frac{1}{10}) = \frac{v_{max}^2}{2} (\frac{3}{10})$ .

$1500 = v_{max}^2 (\frac{3}{20}) \implies v_{max}^2 = 10000 \implies v_{max} = 100 \text{ m/s}$ .

Find Time:

$T = 100 (\frac{1}{5} + \frac{1}{10}) = 100 (0.2 + 0.1) = 100(0.3) = 30 \text{ s}$ .

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