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A motor car of mass 2000 kg is moving with a certain velocity. It is brought to rest by the application of brakes, within a distance of 20 m when the average resistance being offered to it is 5000 N. What was the velocity of the motor car?
Answer:
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Mass of the Car [tex]=2000kg[/tex]
Since, the Force acting on the car as the resistance is in the opposite direction,
Force being offered = [tex]-5000N[/tex]
By Using,
[tex]F=ma[/tex]
[tex]-5000=(2000)a[/tex]
[tex]a=-\dfrac{5}{2}[/tex]
Let’s assume the Initial Velocity to be [tex]u[/tex]
Final Velocity [tex]=0[/tex]
Distance covered [tex]=20m[/tex]
By Using Kinematical Equation of Motion,
[tex] {v}^{2} – {u}^{2} = 2as[/tex]
[tex] {0}^{2} – {u}^{2} = \cancel{2} \left( – \dfrac{5}{ \cancel{2}} \right) (20)[/tex]
[tex] \cancel{ – } {u}^{2} = \cancel{ – } 100[/tex]
[tex]u = \sqrt{100} [/tex]
[tex] \boxed{u = 10m {s}^{ – 1} }[/tex]