5 16 Calculate the maximum deceleration of a car that is heading down a 12.5 slo

Question

5 16

Calculate the maximum deceleration of a car that is heading down a 12.5 slope (one that makes an angle of 12.5 with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the static coefficient of friction is involved--that is, the tires are not allowed to slip during the deceleration.

(a) on dry concrete
1      m/s2
(b) on wet concrete
2      m/s2
(c) on ice, assuming that s = 0.100, the same as for shoes on ice
3      m/s2

Explanation / Answer

1.0 static (dry)
0.30 static (wet)

1)

mgsinx - fmgcosx = ma

9.8*sin12.5 - 1.0*9.8*cos12.5 = a

a = 2.121 - 9.704

a = -7.446 m/s2.

so the deceleration d = 7.446 m/s2

2)

mgsinx - fmgcosx = ma

9.8*sin12.5 - 0.3*9.8*cos12.5 = a

a = 2.121 - 2.87

a = -0.749 m/s2

so the decelaration d = 0.749 m/s2.

3)

mgsinx - fmgcosx = ma

9.8*sin12.5 - 0.1*9.8*cos12.5 = a

a = 2.121 - 0.956

a = 1.164

so the acceleration is 1.164 m/s2.

the maximum deceleration occurs when the car travels on dry concrete.that is 7.446 m/s2.

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