Stopping Distance And Deceleration

Question

A motorcyclist travelling at 30 m/s applies the brakes and decelerates uniformly at 5 m/s^2 until coming to a complete halt. Over what straight-line distance does the motorcycle travel while braking?

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Braking to rest with constant deceleration is uniformly accelerated motion with a negative acceleration, so the time-independent equation of motion $v^2 = u^2 + 2as$ is the most direct tool because it links the known initial and final speeds to the unknown distance. At the moment the motorcycle stops, the final velocity $v = 0$. With $u = 30	ext{ m/s}$ and $a = -5	ext{ m/s}^2$, the relation becomes $0 = 30^2 + 2(-5)s$, giving $0 = 900 - 10s$, so $s = 90	ext{ m}$. The option 180 m is wrong because it omits the factor of two in the denominator, effectively using $u^2/(a)$. The option 60 m incorrectly applies $u^2/(2 	imes 7.5)$ with a wrong deceleration. The option 45 m halves the correct result, as though the speed were 30 but the deceleration 10. This is the NCERT stopping-distance relation, where braking distance scales with the square of the initial speed. Physically, this quadratic dependence arises because kinetic energy itself scales with the square of speed, and the braking force must dissipate all of that energy over the stopping distance. As a plausibility check, stopping distance growing as the square of speed correctly warns that doubling the speed would quadruple the distance needed to stop.

Stopping Distance And Deceleration

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