An automobile is traveling on a long, straight highway at a steady 76.0 mi/h when the driver sees a wreck 180 m ahead. at that instant, she applies the brakes (ignore reaction time). between her and the wreck are two different surfaces. first there is 130 m of ice, where the deceleration is only 1.20 m/s2 . from then on, it is dry concrete, where the deceleration is a more normal 7.10 m/s2 .

a) The car’s speed just after leaving the icy portion of the road is the first part

We have s = ut +

Here s = 130 m, a = -1.2 , u = 76 mile/hr = 33.975 m/s

After 14.72 seconds cars speed is given by v =u +at

                                                          u = 33.975m/s, a = -1.2

                                                      v = 33.975-1.2*14.72 = 16.31 m/s

   So velocity of car on leaving icy surface = 16.31 m/s

b) Time taken after icy surface

              0 = 16.31 - 7.10*t

              t = 2.23 seconds

         Distance traveled during this time

   Total distance traveled = 130+18.74 = 148.74 m

c) Total time taken = 14.72+2.23=16.95 seconds

explanation:

the masses of the objects and the distance between them

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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