a toy friction car is propelled forward from rest by a girl's hand which causes it to reach a forward velocity of 3.5m/s in 0.4 seconds. it is then rreleased and later stops in 5.6 seconds

Respuesta :

Answer:

a)  Acceleration = [tex]8.75m/s^2[/tex]

b) Slow down acceleration = 0.625[tex]m/s^2[/tex]

c) Car travels 9.8 m

Explanation:

  Acceleration = Change in velocity/Time

a) Change in velocity = 3.5-0 = 3.5 m/s

   Time = 0.4 seconds

    Acceleration = [tex]3.5/0.4=8.75m/s^2[/tex]

b) Change in velocity = 0 - 3.5 = -3.5 m/s

   Time = 5.6 seconds

   Acceleration = [tex]-3.5/5.6=-0.625m/s^2[/tex]

   Slow down acceleration = 0.625[tex]m/s^2[/tex]

c) We have equation of motion, [tex]v^2=u^2+2as[/tex], where u is the initial velocity, u is the final velocity, s is the displacement and a is the acceleration.

 Here v = 0 m/s, u = 3.5 m/s, a = -0.625[tex]m/s^2[/tex]

 Substituting

    [tex]0^2=3.5^2-2*0.625*s\\ \\ s=9.8m[/tex]

  So, car travels 9.8 m