Answer to Question #143432 in Classical Mechanics for Sabrina Farro

Question #143432
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?
1
Expert's answer
2020-11-11T07:48:06-0500

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

tf=SfVf=405=8st_f=\frac{S_f}{V_f}=\frac{40}{5}=8s

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

t=δ+ta+tvt =\delta+t_a+t_v

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

ta Time of uniformly accelerated motiont_a \text{ Time of uniformly accelerated motion}

V=V0+ata; V0=0; ta=Va=72=3.5sV =V_0+at_a;\ V_0=0;\ t_a=\frac{V}{a}=\frac{7}{2}=3.5s

Sa=V0ta+ata22=23.522=12.25mS_a =V_0t_a+\frac{at_a^2}{2}=\frac{2*3.5^2}{2}=12.25m

tv Running time at constant speedt_v\text{ Running time at constant speed}

tv=SSaV=3012.2572.54st_v=\frac{S-S_a}{V}=\frac{30-12.25}{7}\approx2.54s

t=δ+ta+tv=1+3.5+2.54=7.04t =\delta+t_a+t_v= 1+3.5+2.54=7.04

t<tft<t_f

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



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