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David is driving a steady 26.0 m/s when he passes tina, who is sitting in her car at rest. tina begins to accelerate at a steady 2.80 m/s2 at the instant when david passes. what is her speed as she passes him?

Respuesta :

Relative speed of Tina with respect to David is given by

[tex]v_r = v_t - v_d[/tex]

[tex]v_r = 0 - 26[/tex]

[tex]v_r = - 26 m/s[/tex]

now the acceleration of Tina with respect to David

[tex]a_r = a_t - a_d[/tex]

[tex]a_r = 2.80 - 0 [/tex]

[tex]a_r = 2.80 m/s^2[/tex]

now the relative displacement would be zero when Tina cross David

so now we have

[tex]\deta x = 0 = v_r * t + \frac{1}{2} a_r t^2[/tex]

[tex]0 = -26 * t + \frac{1}{2}*2.8*t^2[/tex]

[tex]t = 18.6 s[/tex]

now the speed of Tina at this moment is given as

[tex]v_f = v_i + a * t[/tex]

[tex]v_f = 0 + 2.8 * 18.6[/tex]

[tex]v_f = 52 m/s[/tex]

so the speed will be 52 m/s

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