Question

Stoping distance of a vehicles: when brakes are applied to a moving vehicle,the distance travel before stopping is called stoping distance. It is an important factor for road safety and depends on the initial velocity (vo) and the braking capacity or deceleration,-a that is caused by the braking . Derive an expression for stopping distance of a vehicles in terms of vo and a

Answer 1

Answer:

Given initial velocity (u) = vo

deceleration (a) = -a

final velocity (v) = 0

stopping distance (s) = s

we know that

v = u + at

0 = vo - at

$t = \frac{vo}{a}$

$s = ut + \frac{1}{2} a {t}^{2} \\ s = (vo) \frac{vo}{a} - \frac{1}{2}a {( \frac{vo}{a}) }^{2} \\ s = \frac{ {(vo)}^{2} }{a} - \frac{ {(vo)}^{2} }{2a} \\ s = \frac{ {(vo)}^{2} }{2a}$

hence a expression is derived

Answer 2

$\huge\purple{\mathfrak{answer}}$

initial velocity = u (or v0)

final velocity= v =0

acceleration =-a

stopping distance=S

according to first equation of motion :

v=u+at

0 = u + (-a)t

u = at

°•° t = u/a ...(i)

according to second equation of motion :

S = u + 1/2at^2 ...{ii)

putting the value of (i) in (ii)

S = u×(u/a) + 1/2 ×(-a) × (u/a)^2

S = u^2/a - u^2/2a

S = 2u^2-u^2/2a

°•° S = u^2/2a

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