Answer:
A.
Explanation:
Our values are defined by,
[tex]m=1000kg\\v = 2m/s\\a = 2.7 \angle 45\°[/tex]
We can express also in a vectorial way, then
[tex]v=2\hat{i} \rightarrow[/tex]no values in [tex]\hat{j}[/tex]
Acceleration as follow,
[tex]a= 2.7cos45 \hat{i} + 2.7 sin45 \hat{j}[/tex]
We know that velocity is given by,
[tex]V(t) = v_0+at \rightarrow v_0 = 2[/tex]
We need to calculate for t=3, then
[tex]v(3) = 2\hat{i} +(2.7cos45 \hat{i} + 2.7 sin45 \hat{j})*3[/tex]
[tex]v(3) = (2\hat{i}+3*2.7cos45 \hat{i})+(3*2.7 sin45 \hat{j})[/tex]
[tex]v(3) = (2+3*2.7cos45)\hat{i}+(5.7 sin45 )\hat{j}[/tex]
[tex]v(3) = 7.7\hat{i}+5.7\hat{j}[/tex]
Our mass is 1000Kg, so the momentum is
[tex]P = mv[/tex]
[tex]P= 1000(7.7\hat{i}+5.7\hat{j})[/tex]
[tex]P=7700\hat{i}+5700\hat{j}[/tex]
The possible value for the boat's momentum at t = 3.0 is 7700Î+5700j
This can be defined as the product of the mass of a particle and its velocity.
mass = 1000kg , v = 2m/s , 2.7∠45°
Vectorial representation = 2.7cos45Î + 2.7sin45j
V₍t₎ = v₀ + at = v₀ = 2
V₍₃₎ ₎= 2Î + (2.7cos45Î + 2.7sin45j) ₓ 3
V₍₃₎ = (2 + 3 ₓ 2.7cos45) Î + ( 5.7 sin45) j
V₍₃₎ = 7.7 Î + 5.7j
Momentum = mass × velocity
= 1000 ( 7.7 Î + 5.7j )
= 7700Î+5700j
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