# Physics

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You are driving to the grocery store at 14.8 m/s. You are 140.0 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.440 s and that your car brakes with constant acceleration. You are 133 m from the intersection when you begin to apply the brakes.

a) What acceleration will bring you to rest as you just reach the intersection?
b) How long does it take you to stop?

• Physics -

Ok, during the reaction time, he moves
14.8*.440 m
subtract that from 133 to get stopping distance.

solve for a

Time? Vf=Vi+at
solve for t.

• Physics -

i do not think that is correct because i am getting the incorrect answer.

• Physics -

Kinematics 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

ax = -0.820 m/s^2
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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

Δt = 12.1 s
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Hope that helped- keep those kinematic equations handy (or, memorize them)!