contestada

Two cars are initially moving with speeds vA and vB. The cars are decelerated at the same rate until they come to a stop. If it takes car A four times as far to stop as car B, then how do their initial speeds compare?

Respuesta :

Answer:

[tex]v_a = 2 v_b[/tex]

Explanation:

As we know that the speed of car A and car B is given by

[tex]v_a[/tex] & [tex]v_b[/tex]

now we know that both cars are decelerated by same deceleration and stopped finally

so the distance moved by the car is given by the equations

[tex]v_f^2 - v_i^2 = 2a d[/tex]

[tex]0 - v_a^2 = 2(-a) d_a[/tex]

[tex]d_a = \frac{v_a^2}{2a}[/tex]

similarly we have

[tex]d_b = \frac{v_b^2}{2a}[/tex]

now we know that

[tex]d_a = 4 d_b[/tex]

[tex]\frac{v_a^2}{2a} = 4 \frac{v_b^2}{2a}[/tex]

[tex]v_a = 2 v_b[/tex]

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