Question

A man starts walking at a steady speed of 1 m/s and 6s later his son sets off from the same point in pursuit of him, starting at rest and accelerating at 2/3 m/s/s. How far do they go before they are together?

Answer 1

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:

$x=vt=1*6=6\ m$

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:

$\displaystyle xs=v_o.t_1+\frac{a.t_1^2}{2}$

Since vo=0:

$\displaystyle xs=\frac{a.t_1^2}{2}$

The father will continue with constant speed and travels a distance of:

$xf=v.t_1$

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:

$xs=xf+6$

Substituting the equations of each man:

$\displaystyle \frac{a.t_1^2}{2}=v.t_1+6$

We know a=2/3, v=1:

$\displaystyle \frac{2}{3}\cdot\frac{t_1^2}{2}=t_1+6$

Simplifying:

$\displaystyle \frac{t_1^2}{3}=t_1+6$

Multiply by 3:

$t_1^2=3t_1+18$

Rearranging:

$t_1^2-3t_1-18=0$

Factoring:

$(t_1-6)(t_1+3)=0$

Solving:

$t_1=6 , t_1=-3$

Since time cannot be negative, the only valid solution is

$t_1=6\ s$

The distance traveled by the son in 6 seconds is:

$\displaystyle xs=\frac{2/3\cdot 6^2}{2}$

$xs=12\ m$

Note the father will travel

$xf=1*6=6\ m$

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