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The distance required for a car to come to a stop will vary depending on how fast the car is moving. Suppose that a certain car traveling down the road at a speed of 10 m / s can come to a complete stop within a distance of 20 m . Assuming the road conditions remain the same, what would be the stopping distance required for the same car if it were moving at speeds of 5 m / s , 20 m / s , or 40 m / s ?

Respuesta :

Answer:

5 m

80 m

320 m

Explanation:

[tex]v_{o}[/tex] = Initial speed of the car = 10 ms⁻¹

[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹

[tex]d[/tex] = Stopping distance of the car = 20 m

[tex]a[/tex] = acceleration of the car

On the basis of above data, we can use the kinematics equation

[tex]v_{f}^{2} = v_{o}^{2} + 2 a d\\0^{2} = 10^{2} + 2 (20) a\\a = - 2.5 ms^{-2}[/tex]

[tex]v_{o}[/tex] = Initial speed of the car = 5 ms⁻¹

[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹

[tex]d'[/tex] = Stopping distance of the car

[tex]a[/tex] = acceleration of the car = - 2.5 ms⁻²

On the basis of above data, we can use the kinematics equation

[tex]v_{f}^{2} = v_{o}^{2} + 2 a d'\\0^{2} = 5^{2} + 2 (- 2.5) d'\\d' = 5 m[/tex]

[tex]v_{o}[/tex] = Initial speed of the car = 20 ms⁻¹

[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹

[tex]d''[/tex] = Stopping distance of the car

[tex]a[/tex] = acceleration of the car = - 2.5 ms⁻²

On the basis of above data, we can use the kinematics equation

[tex]v_{f}^{2} = v_{o}^{2} + 2 a d''\\0^{2} = 20^{2} + 2 (- 2.5) d''\\d'' = 80 m[/tex]

[tex]v_{o}[/tex] = Initial speed of the car = 40 ms⁻¹

[tex]v_{f}[/tex] = Final speed of the car = 0 ms⁻¹

[tex]d'''[/tex] = Stopping distance of the car

[tex]a[/tex] = acceleration of the car = - 2.5 ms⁻²

On the basis of above data, we can use the kinematics equation

[tex]v_{f}^{2} = v_{o}^{2} + 2 a d'''\\0^{2} = 40^{2} + 2 (- 2.5) d'''\\d''' = 320 m[/tex]

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