Objects A and B are at the origin. If object A moves to #(5 ,-1 )# and object B moves to #(-7 ,2 )# over #3 s#, what is the relative velocity of object B from the perspective of object A? Assume that all units are denominated in meters.

Answer 1

#vec v_(AB)~=sqrt153/3=4,123m/s#


#vec (BC)=sqrt((5+7)^2+(2+1)^2)#
#vec (BC)=sqrt( 12^2+3^2)#
#vec (BC)=sqrt 144+9=sqrt 153 m#
#vec v_(AB)~=sqrt153/3=4,123m/s#

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Answer 2

The question is about velocity which is a vector, hence you need to consider size and direction.

From A's perspective, B will have moved a distance of 12m in the negative x direction (5--7) and 3m in the positive y direction (2--1).

If you are using i and j as unit vectors in the x and y direction, then the displacement vector relative to B would be -12i + 3j. Hence dividing by 3 gives the velocity vector of -4i+ j.

If you are not using this notation, the total distance using Pythagoras = the square root of (12 squared + 3 squared) = 12.4m. Over 3 seconds, this equates to 4.1m/s (an assumption being that points are not accelerating relative to each other).

In terms of direction, you can work this out using trigonometry by defining an angle. This is easier to see if you plot on a graph, but if you defined the positive y axis as 0 degrees, then the angle would be (270 + tan-1 (3/12)

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Answer 3

To find the relative velocity of object B from the perspective of object A, we need to subtract the velocity of object A from the velocity of object B.

Given: Position of object A at ( (x_A, y_A) = (0, 0) ) Position of object B at ( (x_B, y_B) = (5, -1) ) after 3 seconds

Velocity of object A: ( v_A = \frac{\text{Change in position}}{\text{Change in time}} = \left(\frac{5 - 0}{3}, \frac{-1 - 0}{3}\right) = \left(\frac{5}{3}, -\frac{1}{3}\right) ) m/s

Velocity of object B: ( v_B = \frac{\text{Change in position}}{\text{Change in time}} = \left(\frac{-7 - 0}{3}, \frac{2 - 0}{3}\right) = \left(-\frac{7}{3}, \frac{2}{3}\right) ) m/s

Relative velocity of B with respect to A: [ v_{\text{rel}} = v_B - v_A = \left(-\frac{7}{3} - \frac{5}{3}, \frac{2}{3} + \frac{1}{3}\right) = \left(-\frac{12}{3}, \frac{3}{3}\right) = \left(-4, 1\right) ] m/s

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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