A body is moving along a circular path with variable speed, it has both radial and tangential acceleration.
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Answer:
D₂= 167,21 cm : Magnitude of the second displacement
β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement
Explanation:
We find the x-y components for the given vectors:
i: unit vector in x direction
j:unit vector in y direction
D₁: Displacement Vector 1
D₂: Displacement Vector 2
R= resulta displacement vector
D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j
D₂= -D₂(i)-D₂(j)
R= 131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j
We propose the vector equation for sum of vectors:
D₁+ D₂= R
-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j
-51.99i+D₂x(i)=103.23i
D₂x=103.23+51.99=155.22 cm
+142.83j-D₂y(j) =+80.65j
D₂y=142.83-80.65=62.18 cm
Magnitude and direction of the second displacement
D₂= 167.21 cm
Direction of the second displacement
β= 21.8°
D₂= 167,21 cm : Magnitude of the second displacement
β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement
After 20 s, the motorcycle attains a speed of
and it continues at this speed for the next 40 s. So at 45 s, its speed is 80 m/s.