A driver of a 2250 lb automobile falls asleep and crashes into a cushion at 80 m

Question

A driver of a 2250 lb automobile falls asleep and crashes into a cushion at 80 mph as shown. Assume the automobile strikes squarely the highway crash cushion of the type shown in which the automobile is brought to rest by successively crushing steel barrels. The magnitude F the force required to crush the barrels is shown as a function of the distance the automobile has moved into the cushion. Determine:

(a) distance the automobile will move into the cushion before it comes to rest

(b) the maximum deceleration of the automobile.

(NOTE the distance correlates with the number of barrels)

Explanation / Answer

a)

a = -F/m

a = dv/dt = (dv/dx)(dx/dt) = v (dv/dx)

v dv = a dx

v dv = -(F/m) dx

Integrating between 0<=x<=5 ft,

v^2/2 = -(18*10^3*32.2)/2250 x + C where 0<x<5

At x = 0, v = 80 mph = 80*5280/(60*60) ft/s = 117.3 ft/s

117.3^2/2 = 0 + C

C = 6883.5

Thus, v^2/2 = -(18*10^3*32.2)/2250 x + 6883.5 where 0<x<5

v^2/2 = -257.6*x + 6883.5 where 0<x<5

From this equation, at x = 5 ft, we get v = 105.8 ft/s

Again using, v dv = -(F/m) dx and integrating between 5<=x<=15,

v^2/2 = -(27*10^3*32.2)/2250 x + C

At x = 5 ft, v = 105.8 ft/s.

Thus, 105.8^2/2 = -(27*10^3*32.2)/2250 *5 + C

C = 7528.8

So, v^2/2 = -(27*10^3*32.2)/2250 x + 7528.8

or, v^2/2 = -386.4*x + 7528.8

From this equation, at x = 15 ft, we get v = 58.9 ft/s

Again using, v dv = -(F/m) dx and integrating between 15<=x<=25,

v^2/2 = -(36*10^3*32.2)/2250 x + C

At x = 15 ft, v = 58.9 ft/s.

Thus, 58.9^2/2 = -(36*10^3*32.2)/2250 *15 + C

C = 9462.6

So, v^2/2 = -(36*10^3*32.2)/2250 x + 9462.6

or, v^2/2 = -515.2*x + 9462.6

When v = 0 we get x = 18.37 ft

b)

We have, v^2/2 = -257.6*x + 6883.5 where 0<x<5

and v^2/2 = -386.4*x + 7528.8 where 5<x<15

and v^2/2 = -515.2*x + 9462.6 where 15<x<25.

Differentiating these with respect to x we get,

v(dv/dx) = -257.6 where 0<x<5

and v(dv/dx) = -386.4 where 5<x<15

and v(dv/dx) = -515.2 where 15<x<25.

a = dv/dt = (dv/dx)(dx/dt) = v(dv/dx)

Thus max deceleration = 515.2 ft/s^2

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