Question

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?

Answer 1

Relative speed of Tina with respect to David is given by

`v_r = v_t - v_d`

`v_r = 0 - 26`

`v_r = - 26 m/s`

now the acceleration of Tina with respect to David

`a_r = a_t - a_d`

`a_r = 2.80 - 0`

`a_r = 2.80 m/s^2`

now the relative displacement would be zero when Tina cross David

so now we have

`\deta x = 0 = v_r * t + (1)/(2) a_r t^2`

`0 = -26 * t + (1)/(2)*2.8*t^2`

`t = 18.6 s`

now the speed of Tina at this moment is given as

`v_f = v_i + a * t`

`v_f = 0 + 2.8 * 18.6`

`v_f = 52 m/s`

so the speed will be 52 m/s