1. ## speed

(a) A car traveling at a speed of v can brake to an emergency stop in a distance x. Assuming all other driving conditions are all similar, if traveling speed of the car doubles, the stopping distance will be (1) √2x, or (2) 2x, or (3) 4x: (b) a driver traveling at 40.0 km/h in a school zone can brake to an emergency stop in 3.00 m. What would be the braking distance if the car were traveling at 60.0 km/h.

2. Originally Posted by twistedmexican
(a) A car traveling at a speed of v can brake to an emergency stop in a distance x. Assuming all other driving conditions are all similar, if traveling speed of the car doubles, the stopping distance will be (1) √2x, or (2) 2x, or (3) 4x: (b) a driver traveling at 40.0 km/h in a school zone can brake to an emergency stop in 3.00 m. What would be the braking distance if the car were traveling at 60.0 km/h.
s = ut + 0.5 a t

If $\displaystyle t$ and $\displaystyle a$ stay the same, then i would say that $\displaystyle s$ is directly proportionate to $\displaystyle u$.

Therefore the stopping distance is $\displaystyle 2 \ times \ greater$

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For question b)

$\displaystyle v^2 = u^2 + 2as$

40 km/h = 11.1111111 m/s

$\displaystyle 0 = (11.11111)^2 + 2(a)(3)$

$\displaystyle a = -20,5 m/s^2$

60 km/h = 16,67 m/s

$\displaystyle 0 = (16,67)^2 + 2(-20,5)(s)$
$\displaystyle s = 6,77 m$