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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 .

Respuesta :

Blacklash

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

We have s = ut +[tex]\frac{1}{2} at^2[/tex]

Here s = 130 m, a = -1.2 [tex]m/s^2[/tex], u = 76 mile/hr = 33.975 m/s

             [tex]130 = 0*t - 0.5*(-1.2)*t^2\\ \\ \\ t=14.72 seconds[/tex]

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

                                                          u = 33.975m/s, a = -1.2 [tex]m/s^2[/tex]

                                                      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

                   [tex]v^2 = u^2+2as\\ \\ 0 = 16.31^2 - 2*7.10*s\\ \\ s = 18.74 m[/tex]

   Total distance traveled = 130+18.74 = 148.74 m

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

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