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Two cars start from rest at a red stop light. When the light turns green, both...

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 3.6 m/s2 for 4.4 seconds. It then continues at a constant speed for 8.2 seconds, before applying the brakes such that the car

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part a

From ist equation of motion

           Vf =Vi+at

            Here Vi=0

            Vf =at = 1.8(3.6)

            Vf = 6.48m/s

b)   we know that the velocity after 4sec will be

           V =at =3.6(4.4) = 15.84m/s

              After 4sec time the blue car moves 8.2sec more with constant speed. This simply means the speed of the car after 9.8sec will be 15.84m/s

c) Distance cover during ist 4.4sec

            2as1 =Vf2 -Vi2

              s1= Vf2 -VI2 / 2a = 14.42 -02 / 2(3.6) = 28.8m

Distance covered during next 8.2sec

               S2=Vt =15.84(8.2) = 129.88m

Adding the distances

               S =s1+s2

                  =28.8+129.88

                  =158.68m

d) a=?

            Distance cover during deacceleration is 208.3-158.68 = 49.61m

                          From third equation of motion

                            a = Vf2 -VI2 / 2S        ( In this case Vf=0m/s, Vi =15.84m/s)

                              a = -15.842 /2(49.61) = -2.55m/s2

       e)       Time taken during deacceleration is

                  Vf =Vi+at

                   t =- Vi/a =- 15.84/ -2.55 = 6.2sec

Total time = 4.4+8.2+6.2

               = 18.8sec

f ) The acceleration of yellow car will be

             s=vit+0.5at2 = 0+ 0.5at2

              a = s/o.5t2 = 208.3 / 0.5(18.8) = 22.15m/s2

            

       

                 

          

genius_generous answered 4 months ago
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