Question

A car moves 40m due east and turns
towards north and moves 30m then turns
45° east of north and moves 20✔2m. The
net displacement of car is ( east is taken
positive x-axis, North as positive y-axis)
1)50î +60j 2)60î +50j
3) 30î +40j 4) 40î +30j​

Answer 1

Explanation:

In that attachment clear understanding of that question answer is given.

so, the correct option is 2 one.

Answer 2

The correct answer is option (2) 60î +50j

GIVEN

Distance of Car in east direction = 40m

Distance of Car in North direction= 30m

Distance of Car in the east of north direction = 20√2m

Angle = 45°

TO FIND

The net displacement of the car.

SOLUTION

We can simply solve the above problem as follows-

Let,

the Distance traveled in the east direction is, AB

Distance traveled in North direction be, BC.

Distance traveled in the east of north direction be CD.

So,

Net displacement = A̅B̅+ B̅C̅ + C̅D̅

So,

the unit vector in the x-axis = AB = 40î

The unit vector in the y-axis = BC = 30ĵ

Now, The resultant vector of CD = CE +CF

20√2 cos45°î + 20√2sin45°ĵ

We know,

cos45° = sin45° = 1/√2

Therefore,

CD = 20î + 20ĵ

Putting the values of A̅B̅, B̅C̅, C̅D̅

net displacement = 40î+30ĵ+ 20î+20ĵ

= 60î+50ĵ

Hence, the net displacement of the car is 60î+50ĵ

Picture for reference.

#Spj2

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