Answer: 1 m/s
Explanation:
We have an object whose position [tex]r[/tex] is given by a vector, where the components X and Y are identified by the unit vectors [tex]i[/tex] and [tex]j[/tex] (where each unit vector is defined to have a magnitude of exactly one):
[tex]r=[2 m + (2 m/s) t] i + [3 m - (1 m/s^{2})t^{2}] j[/tex]
On the other hand, velocity is defined as the variation of the position in time:
[tex]V=\frac{dr}{dt}[/tex]
This means we have to derive [tex]r[/tex]:
[tex]\frac{dr}{dt}=\frac{d}{dt}[2 m + (2 m/s) t] i + \frac{d}{dt}[3 m - (1 m/s^{2})t^{2}] j[/tex]
[tex]\frac{dr}{dt}=(2 m/s) i - (\frac{1}{2} m/s^{2} t) j[/tex] This is the velocity vector
And when [tex]t=2s [/tex] the velocity vector is:
[tex]\frac{dr}{dt}=(2 m/s) i - (\frac{1}{2} m/s^{2} (2 s)) j[/tex]
[tex]\frac{dr}{dt}=2 m/s i - 1m/s j[/tex] This is the velocity vector at 2 seconds
However, the solution is not complete yet, we have to find the module of this velocity vector, which is the speed [tex]S[/tex]:
[tex]S=\sqrt {-1 m/s j + 2 m/s i}[/tex]
[tex]S=\sqrt {1 m/s}[/tex]
Finally:
[tex]S=1 m/s[/tex] This is the speed of the object at 2 seconds
