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Assume that the brakes in your car create a constant deceleration of 2.8 m/s^{2} regardless of how fast you are driving. Suppose you double your driving speed from15 m/s to 30 m/s.

(a) Does the distance needed to stop increase by a factor of two or a factor of four?

(b) Calculate the stopping distance for the initial speed of 15 m/s.

(c) Calculate the stopping distance for the initial speed of 30 m/s.

m

Use part (b) and (c) to verify your answer for part (a).

Added Wed, 09 Sep '15

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Vf = Vo + a*t ==> t=[Vf - Vo] / a.... putting it in 2nd equation

xf = xo + vo*t + 1/2*a*t^2, x0 =0

xf = vo*t + 1/2*a*t^2= v_{0}*[Vf - Vo] / a + 1/2*a*[{Vf - Vo} / a]^2

by these we get

V^{2}=u^{2} +2ax where V=V_{f} , u=V_{0},

(a) the distance needed to stop increase by a factor of four

(b) the stopping distance for the initial speed of 15 m/s.

V^{2}=u^{2} +2ax

0=15^{2} - 2*2.8x

x=225/5.6 =40.1785

(c) the stopping distance for the initial speed of 30 m/s.

is

V^{2}=u^{2} +2ax

0=30^{2} - 2*2.8x

x=900/5.6 =160.71

Using part (b) and (c)

160.71 /40.1785 =4

Added Wed, 09 Sep '15

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