Answer:
The father will walk 6 more meters and the son will walk 12 meters
Explanation:
Uniform Speed and Acceleration
This is a problem where two objects have different types of motion. The father walks at a constant speed and later, his son starts a constant acceleration motion in pursuit of him.
Let's start with the father, whose speed is v=1 m/s during t=6 seconds. He travels a distance:
[tex]x=vt=1*6=6\ m[/tex]
Now the son, starting from rest (vo=0) accelerates at a=2/3 m/s^2. His speed will increase and eventually, he will catch up with his father. Let's suppose it happens at a time t1.
The distance traveled by the son is given by:
[tex]\displaystyle xs=v_o.t_1+\frac{a.t_1^2}{2}[/tex]
Since vo=0:
[tex]\displaystyle xs=\frac{a.t_1^2}{2}[/tex]
The father will continue with constant speed and travels a distance of:
[tex]xf=v.t_1[/tex]
For them to catch up, the distance of the son must be 6 m more than the distance of the father, because of the leading distance he has already taken. Thus:
[tex]xs=xf+6[/tex]
Substituting the equations of each man:
[tex]\displaystyle \frac{a.t_1^2}{2}=v.t_1+6[/tex]
We know a=2/3, v=1:
[tex]\displaystyle \frac{2}{3}\cdot\frac{t_1^2}{2}=t_1+6[/tex]
Simplifying:
[tex]\displaystyle \frac{t_1^2}{3}=t_1+6[/tex]
Multiply by 3:
[tex]t_1^2=3t_1+18[/tex]
Rearranging:
[tex]t_1^2-3t_1-18=0[/tex]
Factoring:
[tex](t_1-6)(t_1+3)=0[/tex]
Solving:
[tex]t_1=6 , t_1=-3[/tex]
Since time cannot be negative, the only valid solution is
[tex]t_1=6\ s[/tex]
The distance traveled by the son in 6 seconds is:
[tex]\displaystyle xs=\frac{2/3\cdot 6^2}{2}[/tex]
[tex]xs=12\ m[/tex]
Note the father will travel
[tex]xf=1*6=6\ m[/tex]
This 6 m plus the 6 m he was ahead of the son, make them meet while walking at 6 seconds.
Answer: The father will walk 6 more meters and the son will walk 12 meters
