Respuesta :
Answer:0.3568 m/s
Explanation:
Given
Velocity of a canoe is 0.4 m/s southeast relative to earth
Representing in vector form
[tex]V_{c}=\left ( 0.4cos45\right )\hat{i}-\left ( 0.4sin45\right )\hat{j}[/tex]
Also Velocity of river [tex]\left ( V_r\right )=0.5\hat{i}[/tex]
thus velocity of canoe relative to the river
[tex]V_{cr}=V_c-V_r[/tex]
[tex]V_{cr}=\frac{0.4}{\sqrt{2}}\hat{i}-\frac{0.4}{\sqrt{2}}\hat{j}-0.5\hat{i}[/tex]
[tex]V_{cr}=\left ( \frac{0.4\sqrt{2}-1}{2}\right )\hat{i}-\left ( \frac{0.4\sqrt{2}}{2}\right )\hat{j}[/tex]
magnitude of Velocity[tex]=\sqrt{\left ( \frac{0.4\sqrt{2}-1}{2}\right )^2+\left ( \frac{0.4\sqrt{2}}{2}\right )}[/tex]
[tex]V_{net}=0.3568 m/s[/tex]
[tex]tan\theta =\frac{\frac{0.4\sqrt{2}}{2}}{\frac{0.4\sqrt{2}-1}{2}}[/tex]
[tex]\theta =52.47 ^{\circ}[/tex] with east direction in clockwise direction
Velocity of a object is the ratio of distance traveled by the object with the time taken.
- a) The magnitude of the velocity of canoe relative to the river is 0.36 m/s.
- b) The direction angle of the canoe is 52 degree clockwise in east direction.
What is velocity of a object?
Velocity of a object is the ratio of distance traveled by the object with the time taken.
Given information-
The velocity of the canoe is 0.40 m/s
The speed of the river is 0.50 m/s towards east relative to the earth.
The horizontal velocity of the canoe can be given as ,
[tex]v_x=0.5-v_0\cos \theta\\[/tex]
Put the values,
[tex]v_x=0.5-0.4\cos (45^o)\\v_x=0.22\rm m/s[/tex]
Thus the horizontal velocity of the canoe is 0.55 m/s.
The vertical velocity of the canoe can be given as ,
[tex]v_x=v_0\sin\theta[/tex]
Put the values,
[tex]v_x=0.4\sin(45^o)\\v_x=0.28\rm m/s[/tex]
Thus the vertical velocity of the canoe is 0.28 m/s.
Thus the magnitude of the velocity of canoe relative to the river can be given as,
[tex]v_r=\sqrt{v_x^2+v_y^2}\\v_r=\sqrt{0.22^2+0.28^2}\\v_r=0.36\\[/tex]
Thus the magnitude of the velocity of canoe relative to the river is 0.36 m/s.
The direction of the river is,
[tex]\theta=\tan ^{-1}\dfrac{v_y}{v_x} \\\theta=\tan ^{-1}\dfrac{0.28}{0.22} \\\theta=52^o[/tex]
Thus the direction angle of the canoe is 52 degree clockwise in east direction.
Hence,
- a) The magnitude of the velocity of canoe relative to the river is 0.36 m/s.
- b) The direction angle of the canoe is 52 degree clockwise in east direction.
Learn more about the velocity here;
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