Time at which two bodies with u.r.m. will meet

Data

• Speed of first body: V₁ = 20 m/s
• Speed of second body: V₂ = 24 m/s
• Difference in time of departure between the two bodies ∆t = 15 s

Preliminary considerations

• Two bodies begin their motion at different times. The time difference between them is 15 s
• Keep in mind that we can use the sign convention in rectilinear motion that allows us to use scalar magnitudes instead of vectors to describe the motion

Resolution

The expression that allows us to determine the position of each body as a function of speed and time is:

If t₁ is the time that the first body is in motion, then for the first body we get:

$$x_1=x_{01}+v_1\cdot t_1$$ 

t₂ is the time the second body is in motion, then for the second body we get:

$$x_2=x_{02}+v_2\cdot t_2$$ 

Both bodies depart from the same point, therefore em>x₀₁ = x₀₂ = 0, but they do so at different times: t₂  = t₁ - 15 s. Additionally, when they are in the same position x₁ = x₂, resulting in:

$$\begin{gathered} v_1·t_1=v_2·t_2\Rightarrow v_1·t_1=v_2·(t_1-15)\Rightarrow v_1·t_1-v_2·t_1=-15·v_2\Rightarrow \\ \Rightarrow t_1=\frac{-15·v_2}{v_1-v_2}=\frac{-15·24}{20-24}=\boxed{90 s} \end{gathered}$$ 

That is, they meet when the first body has been moving for 90 seconds and the second 90 - 15 = 75 s. To determine where they are when they meet, we simply replace the value of the time into the motion equation of the first body:

$$x_1=x_{01}+v_1\cdot t_1=0+20·90=\boxed{1800 m}$$

Notice that in this exercise, we have used the same values as in this other, but you are asked different magnitudes. As it can be expected, regardless of the unknown variables, the behavior of a body in u.r.m. should be the same.