Question #188524

What is the final velocity of a 1200 kilogram car starting from rest after 50 meters when 4500 Newton's of force is applied?


1
Expert's answer
2021-05-04T12:09:10-0400

According to the second Newton's law, the acceleration of the car is:


a=Fma = \dfrac{F}{m}

where F=4500NF = 4500N is the total force applied to the car and m=1200kgm = 1200kg is the mass of the car.

According to the kinematic equation, the distance covered under a constant acceleration is:


d=v0t+at22=at22d =v_0t + \dfrac{at^2}{2} = \dfrac{at^2}{2}

where v0=0v_0 = 0 is the initial speed of the car (equal to 0 since the car starts from rest) and tt is the time.

Since d=50md = 50m, we can find the time:


t=2da=2dmFt = \sqrt{\dfrac{2d}{a}} = \sqrt{\dfrac{2dm}{F}} \\

The velocity of the car at this moment of time (if it starts from rest) is given by the following kinematic equation:


v=v0+at=atv =v_0 + at = at

Substituting the expressions for aa and tt, obtain:


v=Fm2dmF=2dFmv=250450012006.1m/sv = \dfrac{F}{m}\cdot \sqrt{\dfrac{2dm}{F}} = \sqrt{\dfrac{2dF}{m}}\\ v = \sqrt{\dfrac{2\cdot 50\cdot 4500}{1200}} \approx 6.1 m/s

Answer. 6.1 m/s.


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