Motion at constant acceleration
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# Motion at constant acceleration

[From: ] [author: ] [Date: 11-05-18] [Hit: ]
If you convert from km/hr to m/s, you will be dealing with more reasonable numbers.(85 km/hr) * (10^3 m/km) * (1/3600 hr/s) = 850/36 m/s ≅ 23.If acceleration is constant,where v_f is the final velocity (0 m/s), v_i is the initial velocity (23.......
A car traveling 85km/hr strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80 m. What is the average acceleration of the driver during the collision?
I don't know where to begin with this problem if someone could show me how to do this step by step I would really appreciate it!

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My first thought is to convert the velocity from units of km/hr to units of m/s, just because m/s is the normal velocity measurement in physics. In any case, you must at minimum either convert the car velocity into m/hr or the stopping distance to km. If you convert from km/hr to m/s, you will be dealing with more reasonable numbers.

(85 km/hr) * (10^3 m/km) * (1/3600 hr/s) = 850/36 m/s ≅ 23.61 m/s

If acceleration is constant, then this equation is true:

(v_f)^2 = (v_i)^2 + 2aΔs

where v_f is the final velocity (0 m/s), v_i is the initial velocity (23.61 m/s), a is the acceleration and Δs is the stopping distance. Solving for a:

2aΔs = (v_f)^2 - (v_i)^2

a = ((v_f)^2 - (v_i)^2) / 2Δs

a = -(23.61 m/s)^2 / 2(0.8 m)

a = -557.5 m^2/s^2 / 1.6 m = -348.4 m/s^2

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Use the equation of motion 2aS = v^2 - u^2
Initial velocity ( u ) = 85 km/hr = 85 * (5/18) m/sec = 23.611
final velocity ( v ) = 0 m/sec
Displacement ( S ) = 0.80 m

2a * 0.80 = 0 - 557.484
a = - 348

negative sign implies that this is case of retardation.
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