Question

A toy cart on the floor moves 10m at 37o north of east and then 15m at 53o south of east and finally moves ‘d’ meters at 60o North of east. If the net displacement is towards east, find ‘d’ (take root3 as 1.73)

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Answer 1

Given:

A toy cart on the floor moves 10m at 37° north of east and then 15m at 53° south of east and finally moves ‘d’ meters at 60° North of east. The net displacement is towards east.

To find:

Value of d.

Calculation:

Considering Actual Cartesian Graph:

Let 10 m towards 37° NE be denoted vector a:

$\vec {a} = 10 \cos(37 \degree) \hat{i} + 10 \sin(37 \degree) \hat{j}$

$= > \vec {a} = 8 \hat{i} + 6 \hat{j}$

Similarly , 15 m towards 53° SE be denoted by vector b.

$\vec {b} = 15 \cos(53 \degree) \hat{i} - 15 \sin(53 \degree) \hat{j}$

$= > \vec {b} = 9 \hat{i} - 12 \hat{j}$

Similarly d metres towards 60° NE be denoted be vector c.

$\vec {c} = d \cos(60 \degree) \hat{i} + d \sin(60 \degree) \hat{j}$

$\vec {c} = \dfrac{d}{2} \hat{i} + \dfrac{ \sqrt{3}d }{2} \hat{j}$

So, resultant Displacement will be :

$\vec {d_{net}} =(8 + 9 + \dfrac{d}{2} ) \hat{i} + (6 - 12 + \dfrac{ \sqrt{3}d }{2} ) \hat{j}$

$= > \vec {d_{net}} =(17 + \dfrac{d}{2} ) \hat{i} + ( \dfrac{ \sqrt{3}d }{2} - 6 ) \hat{j}$

Now, as per question , net Displacement is along X axis ; so Y component value should be equal to zero.

$\therefore \: \dfrac{ \sqrt{3}d }{2} - 6 = 0$

$= > \: \dfrac{ \sqrt{3}d }{2} = 6$

$= > \sqrt{3} d = 12$

$= > d = \dfrac{12}{ \sqrt{3} }$

$= > d = \dfrac{12}{ 1.73 }$

$= > d = 6.936 \: m$

So , final answer is :

$\boxed{ \red{ \bold{ \huge{ d = 6.936 \: m}}}}$