Answer:
\[\upsilon
(t)=2t\,\hat{i}~+\text{ }{{t}^{2}}\,\hat{j}\]
\[a(t)=2\hat{i}+2t\,\hat{j}\]
Momentum
at \[t=2s,=mv\left( t \right){{|}_{t\text{ }=\text{ }2}}\]
\[=2\times
[4\hat{i}~+4\hat{j}]=(8\hat{i}~+8\hat{j})\text{ }kg\text{ }m{{s}^{1}}\]
\[a(t){{|}_{t\text{
}=\text{ }2}}=(2\hat{i}+4\hat{j})\]
\ \[F=ma\,{{(t)}_{t\text{ }=\text{
}2}}=(4~\hat{i}+8\hat{j})\text{ }N\]
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