Posted by Emily on Wednesday, October 6, 2010 at 3:10pm.
Ok, 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.
i do not think that is correct because i am getting the incorrect answer.
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
Final answer (significant figures):
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
Final answer (significant figures):
Δt = 12.1 s
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Hope that helped- keep those kinematic equations handy (or, memorize them)!