speed

Quote:

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.:eek::eek:

s = ut + 0.5 a t

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

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

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For question b)

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

40 km/h = 11.1111111 m/s

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

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

60 km/h = 16,67 m/s

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