Question #166189

To go from the student hostels to the university library, a student must cover 650m in a direction 47o south of west. To go from the football ground to the library, the student must move straight south for 420m. How far and in what direction must the student move to go from hostels to the football ground?



1
Expert's answer
2021-03-08T08:32:39-0500

As we know from the condition of the question, to go from the student hostels to the university library, a student must cover 650 m in a direction 4747^{\circ}south of west. To go from the football ground to the library, the student must move straight south for 420 m. So, if the student will go from the library straight north for 420 m, he will go to the football ground. Then, his resultant displacement will be the displacement that he must move to go from hostels to the football ground. Let's find the resultant displacement of the student:


dx=650 mcos(180+47)+420 mcos90=443.3 m,d_x=650\ m\cdot cos(180^{\circ}+47^{\circ})+420\ m\cdot cos90^{\circ}=-443.3\ m,dy=650 msin(180+47)+420 msin90=55.38 m.d_y=650\ m\cdot sin(180^{\circ}+47^{\circ})+420\ m\cdot sin90^{\circ}=-55.38\ m.


Then, the magnitude of the resultant displacement can be found from the Pythagorean theorem:


d=dx2+dy2=(443.3 m)2+(55.38 m)2=446.74 m.d=\sqrt{d_x^2+d_y^2}=\sqrt{(-443.3\ m)^2+(-55.38\ m)^2}=446.74\ m.

We can find the angle as follows:


θ=sin1(dyd)=sin1(55.38 m446.74 m)=7.12.\theta=sin^{-1}(\dfrac{d_y}{d})=sin^{-1}(\dfrac{-55.38\ m}{446.74\ m})=-7.12^{\circ}.


The sign minus means that the resultant displacement has direction 7.12S of W7.12^{\circ} S\ of\ W.

Therefore, the student must move 446.74 m in a direction of 7.12S of W7.12^{\circ} S\ of\ W to go from hostels to the football ground.


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