Answer to Question #143432 in Classical Mechanics for Sabrina Farro

Question

You and a friend are standing far from a door and meet each other’s eyes. Your friend starts running towards the door with a constant speed of 5 m/s while starting 40 meters away from the door. After 1 second passes you become determined not to lose the race to the door and begin running. Starting from rest you accelerate at 2 m/s2 up to your top speed of 7 m/s. You begin your run 30 meters from the door.
Given these parameters, who reaches the door first, you or your friend?

Answer 1

$\text{The time it takes for a friend to reach the door:}$

$t_f=\frac{S_f}{V_f}=\frac{40}{5}=8s$

$\text{The time it takes you to reach the door}:$

$t =\delta+t_a+t_v$

$\text{where } \delta=1s\text{ Time difference between you and your friend start to move}$

$t_a \text{ Time of uniformly accelerated motion}$

$V =V_0+at_a;\ V_0=0;\ t_a=\frac{V}{a}=\frac{7}{2}=3.5s$

$S_a =V_0t_a+\frac{at_a^2}{2}=\frac{2*3.5^2}{2}=12.25m$

$t_v\text{ Running time at constant speed}$

$t_v=\frac{S-S_a}{V}=\frac{30-12.25}{7}\approx2.54s$

$t =\delta+t_a+t_v= 1+3.5+2.54=7.04$

$t<t_f$

Answer: You run the distance to the door before your friend