Question

David is driving a steady 26.0 m/sm/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.50 m/s2m/s2 at the instant when David passes. a. How far does Tina drive before passing David?b. What is her speed as she passes him?

Answer 1

Answer:

a. $x=540.8m$

b. $v_f=52\frac{m}{s}$

Explanation:

a. David is moving with constant speed. While Tina is accelerating, we use the equations of uniformly accelerated motion.

For David we have:

$x=vt$

For Tina:

$x=v_0t+\frac{at^2}{2}\\v_0=0\\x=\frac{at^2}{2}$

They both travel the same distance from the moment David passes her until she passes David:

$vt=\frac{at^2}{2}\\t=\frac{2v}{a}\\t=\frac{2(26\frac{m}{s})}{2.5\frac{m}{s^2}}\\t=20.8s$

Knowing this, we can calculate the distance:

$x=vt\\x=26\frac{m}{s}(20.8s)\\x=540.8m$

b. To calculate her speed when she passes him, we use:

$v_f=v_0+at\\v_f=at\\v_f=(2.5\frac{m}{s^2})20.8s\\v_f=52\frac{m}{s}$