[tex]Step 11 of 4\textbf{Given :}Given :a=5\; \mathrm{m \cdot s^{-2}}a=5m⋅s −2 t=30 \;\mathrm{s}t=30sx=10 \;\mathrm{ km }=10^{4} \;\mathrm{ m }x=10km=10 4 m.\textbf{Required:}Required:v_{i}v i .v_{f}v f .[/tex]
[tex]According to laws of motion the initial velocity v_{i}v i is given by\boxed{x=v_{i}t+ \dfrac{1}{2} at^{2}}\;\;\;\;\;(1)x=v i t+ 21 at 2 (1)substitute and get that10^{4}=v_{i} (30)+.5 (5)(30)10 4 =v i (30)+.5(5)(30)Simplify and getv_{i}=\boxed{258.3\; \mathrm{m/s}}v i = 258.3m/s[/tex]
[tex]According to motion equations the final velocity v_{f}v f is given byv_{f}=v_{i}+at \;\;\;\;\;\;\;(2)v f =v i +at(2)Substituting and get thatv_{f}=258.3+(30)(5)v f =258.3+(30)(5)Simplifying and get thatv_{f}=\boxed{408.3 \;\mathrm{m/s}}v f = 408.3m/s[/tex]
[tex]Result4 of 4v_{i}=258.3\; \mathrm{m/s}v i =258.3m/sv_{f}=408.3\; \mathrm{m/s}v f =408.3m/s[/tex]
It is the rate at which an object's location changes in relation to time and a frame of reference. The action of traveling fast in one direction is all that velocity is, despite its sophisticated appearance
The concept of velocity calls for both magnitude (speed) and direction because it is a vector quantity. Its SI equivalent is the m / sec (ms-1). A body is considered to be accelerating if the direction or magnitude of its velocity changes.
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