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December 20, 2014

December 20, 2014

Posted by **Emily** on Wednesday, October 6, 2010 at 3:10pm.

a) What acceleration will bring you to rest as you just reach the intersection?

b) How long does it take you to stop?

- Physics -
**bobpursley**, Wednesday, October 6, 2010 at 4:36pmOk, during the reaction time, he moves

14.8*.440 m

subtract that from 133 to get stopping distance.

Vf^2=Vi^2+2ad

solve for a

Time? Vf=Vi+at

solve for t.

- Physics -
**Emily**, Wednesday, October 6, 2010 at 4:49pmi do not think that is correct because i am getting the incorrect answer.

- Physics -
**Steven**, Wednesday, October 6, 2010 at 4:52pmKinematics Equations:

Vx1 = Vx0 + axΔt

(Vx1)^2 = (Vx0)^2 + 2axΔx

Δx = Vx0Δt + ½ ax(Δt^2)

Had the distance from the intersection when braking begins not been given:

Remaining distance = 140.0 m

Current Velocity = 14.8 m/s

Reaction time = 0.440 s

Find distance from intersection when breaking begins:

140.0 m - [14.8 m/s (0.440 s)] = 133.488

= 133. 488 m.

Find what acceleration will bring you to rest as you just reach the intersection:

Use the second kinematics equation, rearrange the variables to get "ax" by itself on one side:

(Vx1)^2 = (Vx0)^2 + 2axΔx

= V(Vx1)^2 - (Vx0)^2 = 2axΔx

= [1/(2Δx)] [(Vx1)^2 - (Vx0)^2] = ax

So: ax = [1/(2Δx)] [(Vx1)^2 - (Vx0)^2]

Now plug in quantities for the known variables:

ax = Brake acceleration = ? m/s^2

Δx = Distance to stop = 133.488 m

Vx1 = Final Velocity = 0.00 m/s

Vx0 = Initial Velocity = 14.8 m/s

ax = {1/[2(133.488 m)]} {[0.00 m/s]^2 -[14.8 m/s]^2}

ax = -0.82044828 m/s^2

Final answer (significant figures):

ax = -0.820 m/s^2

--------------------------------------

Use the first kinematics equation to determine how long it will take you to stop:

Vx1 = Vx0 + axΔt

You can rearrange the equation to get Δt by itself, or just plug in what you know:

0.00 m/s = 14.8 m/s + [(-0.82044828 m/s^2)(Δt)

-14.8 m/s = (-0.82044828 m/s^2)(Δt)

Δt = (-14.8 m/s)/(-0.82044828)

Δt = 12.14263454

Final answer (significant figures):

Δt = 12.1 s

------------------------------------

Hope that helped- keep those kinematic equations handy (or, memorize them)!

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