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

$\deta x = 0 = v_r * t + \frac{1}{2} a_r t^2$

$0 = -26 * t + \frac{1}{2}*2.8*t^2$

$t = 18.6 s$

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