Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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?

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?

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
6 years ago
Assumptions:
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.
236-301_image.PNG
(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
236-2485_1111.PNG
Substitute the given values of r1 and r2 in the equation above to have
236-444_comb.PNG
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:
236-1165_delta.PNG
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 positivex\ axis\ be\ \o . Therefore the angle can be defined in terms of the component of displacement vector as:
236-615_11.PNG
Substitute the values of the components rx and ry from above, you have

236-1632_45.PNG
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:
236-596_delta2.PNG
Substitute the value \Delta tof from above and \Delta \overrightarrow{r}from equation (1) into equation (2) to obtain the average velocity vector as:
236-202_v.PNG
Therefore the average velocity vector of the plane is0.22 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 .

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free