# A plane flies 410 mi east from city A to city B in 45 min and then 820 mi south from city B to city C in 1 h 30 min. (a) What are the magnitude and direction of the displacement vector that represents the total trip? What are (b) the average velocity vector and (c) the average speed for the trip?

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A plane flies 410 mi east from city A to city B in 45 min and then 820 mi south from city ย ย B to city C in 1 h 30 min. (a)What are the magnitude and direction of the displacement vector that represents the total trip? What are (b) the average velocity vector and (c) the average speed for the trip?

## 1 Answers

Let us assume that the motion of the plane is described using position vectora , and respectively. Position vector describes the motion when the plane moves from city A to city B, position vector (defined relative to city B) describes the motion when it moves from city B to city C.

We assume that the unit vector points along the positive axis while the unit vector points along the y axis.

Also assume that the displacement vector is defined by vector such that the horizontal and vertical components of vector is r

_{x}and r

_{y}respectively.

Given:

Magnitude of vector mi., .

Time taken to travel from city A to city B,t1 = 45 min. .

Magnitude of vector ,

Time taken to travel from city B to city C, t

_{2}= 1 h 30 min.

The figure below shows the vector diagram for the motion of the plane.

(a) The position vector relative to city A can be defined as:

Vector defined the motion of plane from city A to city B, along east; therefore there is no component of the vector along unit vector which points north.

Similarly position vector relative to city B is given as:

The negative sign in the vertical component accounts for the fact that the plane moves southward, opposite to the direction of unit vector .

Therefore the displacement vector (say ) of the plane can be defined as:

Substitute the vectors calculated above

Substitute the given values of r

_{1}and r

_{2}in the equation above to have

Therefore the displacement vector of the plane is 410 mi

If displacement vector is defined as:

Then the value of its components from can be derived by comparing them with the vector calculated above. On comparison, you have

The magnitude of displacement vector is the given in terms of its vertical and horizontal components as:

Rounding off to two significant figures, we have

Therefore the magnitude of the displacement vector is 920 mi.

Assume that the angle subtended by the displacement vector with respect to the positive . Therefore the angle can be defined in terms of the component of displacement vector as:

Substitute the values of the components r

_{x}and r

_{y}from above, you have

The negative sign shows that the angle is measured clockwise with respect to positive x axis.

Therefore the displacement vector is directed at 63.4° , measured clockwise from positive axis.

(b) The average velocity vector (say )is given as:

…… (2)

The elapse time () is the time taken by the plane to go from city A to city B and then to city C.

Therefore

It is important to note that when we started observing the motion of the plane, was zero. However as the plane moved from city A to B and then to C, the change of the time interval increased such that the time taken by the object to move from city A to city C is t

_{1}+ t

_{2}.

Convert the time t

_{2}in minutes and then add to t

_{1}as:

t

_{2}= 60 x 30 min

= 1800 min

Therefore elapsed time () is:

Substitute the value of from above and from equation (1) into equation (2) to obtain the average velocity vector as:

Therefore the average velocity vector of the plane is .

(c) The average speed of the plane is defined as:

The total distance travelled by the plane is the sum of the magnitude r

_{1}and r

_{2}whereas the total time taken by the plane is equal to its elapse time calculated above.

Therefore the average speed is:

Substitute the given values of r

_{1}, r

_{2}and the calculated value of in the equation above

= 0.66 mi/ min

Therefore the average speed of the plane for the trip is 0.66 mi/min .