Answer:
Explanation:
I will assume the equation reads:
v = 8t²î + 5tĵ
The velocity v is the time derivative of the position x.
[tex]x = \int\limits^t_0 {v} \, dt = \int\limits^t_0 {8t^{2}\hat i + 5t\hat j} \, dt = \frac{8}{3} t^{3} \hat i + \frac{5}{2}t^{2}\hat j |^t_0 = \frac{8}{3} t^{3} \hat i + \frac{5}{2}t^{2}\hat j - \frac{8}{3} \hat i - \frac{5}{2} \hat j\\ x = \frac{8}{3} (t^{3} - 1 )\hat i + \frac{5}{2} (t^{2} - 1 )\hat j[/tex]
The position as a function of time is expressed as [tex]x(t)=\frac{8t^3}{3}i +\frac{5t^2}{2}j[/tex]
Given the velocity function;
[tex]v(t) = 8t^2i + 5tj[/tex]
v(t) is the velocity at a time t
t is the time taken:
To get the position as a function of time, we will have:
[tex]x(t)=\int\limits^t_0 {(8t^2i+5tj)} \, dt\\x(t)=\frac{8t^3}{3}i +\frac{5t^2}{2}j \left \{ {t} \atop 0}} \right.[/tex]
Substitute the given limits into the function'
[tex]x(t)=\frac{8t^3}{3}i +\frac{5t^2}{2}j - (\frac{8(0)^3}{3}i +\frac{5(0)^2}{2}j) \\x(t)=\frac{8t^3}{3}i +\frac{5t^2}{2}j[/tex]
Hence the position as a function of time is expressed as [tex]x(t)=\frac{8t^3}{3}i +\frac{5t^2}{2}j[/tex]
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