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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 B to city C in 1 h 30 min. (a)

6 years ago

Let us assume that the motion of the plane is described using position vectora [\overrightarrow{r_{1}}] , and [\overrightarrow{r_{2}}] respectively. Position vector [\overrightarrow{r_{1}}] describes the motion when the plane moves from city A to city B, position vector [\overrightarrow{r_{2}}] (defined relative to city B) describes the motion when it moves from city B to city C.

We assume that the unit vector [\widehat{i}] points along the positive axis while the unit vector [\widehat{j}] points along the y axis.

Also assume that the displacement vector [\Delta \overrightarrow{r}] is defined by vector such that the horizontal and vertical components of vector is rx and ry respectively.

Given:

Magnitude of vector [\overrightarrow{r_{1}} , r_{1} = 410] mi., .

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

Magnitude of vector [\overrightarrow{r_{2}} , r_{2} = 820 mi] ,

Time taken to travel from city B to city C, t2 = 1 h 30 min.

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

(a) The position vector [\overrightarrow{r}] relative to city A can be defined as:

[\overrightarrow{r_{1}} = \overrightarrow{r_{1}}\widehat{i}]

Vector [\overrightarrow{r_{1}}] defined the motion of plane from city A to city B, along east; therefore there is no component of the vector along unit vector [\widehat{j}] which points north.

Similarly position vector [\overrightarrow{r_{2}}] relative to city B is given as:

[\overrightarrow{r_{2}} = -r_{2}\widehat{j}]

The negative sign in the vertical component accounts for the fact that the plane moves southward, opposite to the direction of unit vector [\widehat{j}] .

Therefore the displacement vector (say [\Delta \overrightarrow{r}] ) of the plane can be defined as:

[\Delta \overrightarrow{r} = \overrightarrow{r_{2}} - \overrightarrow{r_{1}}]

Substitute the vectors [\overrightarrow{r_{1}} and \overrightarrow{r_{2}}] calculated above

Substitute the given values of r1 and r2 in the equation above to have

Therefore the displacement vector of the plane is 410 mi [\widehat{i} - 820 mi \widehat{j}.]

If displacement vector [\Delta \overrightarrow{r}] is defined as:

[\Delta \overrightarrow{r} = r_{x} \widehat{i} + r_{y} \widehat{j}]

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

[r_{x} = 410 mi]

[r_{y} = -820 mi]

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

[|\Delta \overrightarrow{r}| = 920 mi]

Therefore the magnitude of the displacement vector is 920 mi.

Assume that the angle subtended by the displacement vector [\Delta \overrightarrow{r}] with respect to the positive [x\ axis\ be\ \o] . Therefore the angle can be defined in terms of the component of displacement vector as:

Substitute the values of the components rx and ry 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 [\overrightarrow{v_{av}}] )is given as:

[\overrightarrow{v}_{av}] [= \frac{\Delta \overrightarrow{r}}{\Delta t}] …… (2)

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

Therefore [\Delta t = t_{1} + t_{2}]

It is important to note that when we started observing the motion of the plane, [\Delta t] 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 t1 + t2.

Convert the time t2 in minutes and then add to t1 as:

t2 = 60 x 30 min

= 1800 min

Therefore elapsed time ( [\Delta t] ) is:

Substitute the value [\Delta t] of from above and [\Delta \overrightarrow{r}] from equation (1) into equation (2) to obtain the average velocity vector as:

Therefore the average velocity vector of the plane is [0.22 mi/min\widehat{i}-0.44 mi/min \widehat{j}] .

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

[average\ speed = \frac{total\ distance\ travelled}{total\ time \ taken}]

The total distance travelled by the plane is the sum of the magnitude r1 and r2 whereas the total time taken by the plane is equal to its elapse time [\Delta t] calculated above.

Therefore the average speed is:

[average\ speed = \frac{r_{1}+r_{2}}{\Delta t}]

Substitute the given values of r1, r2 and the calculated value of in the equation above

[average\ speed = \frac{410\ mi + 820\ mi}{1845\ min}]

= 0.66 mi/ min

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

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