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Sagot :
Given,
The velocity of the objects, v₁=50.0 m/s and v₂=-25.0 m/s
The initial positions of the objects, x₁_₀=0.00 m and x₂_₀=500 m
The relative velocity of the objects is given by,
[tex]v=v_1-v_2[/tex]On substituting the known values,
[tex]\begin{gathered} v=50.0-(-25.0) \\ =75.0\text{ m/s} \end{gathered}[/tex]The total distance between the objects is,
[tex]\begin{gathered} d=x_{10}+x_{20} \\ =0+500 \\ =500\text{ m} \end{gathered}[/tex](a)
As the two objects are traveling towards each other in a straight line, they will intercept.
The time it takes for the objects to intercept is given by,
[tex]t=\frac{d}{v}[/tex]On substituting the know values,
[tex]\begin{gathered} t=\frac{500}{75} \\ =6.67\text{ s} \end{gathered}[/tex]As t is the time it takes for the objects to intercept, the distance covered by the objects in this time will give us the position x_f where these two objects intercept.
As object 1 starts from the origin, the distance traveled by this object is equal to x_f
The distance traveled by object 1 in that time is,
[tex]x_f=d_1=v_1t[/tex]On substituting the known values,
[tex]\begin{gathered} x_f_{}=50\times6.67 \\ =333.5m\text{ } \end{gathered}[/tex]Thus the objects will intercept at the point x_f=333.5 m from the origin.
(b)
As calculated in part a, the time it takes for the objects to intercept is t=6.67 s
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