Physics

4.10

Since the motorist taking the 60° turn after every 500m, It will form a perfect hexagon of side 500m after 6 turns.

Suppose the motorist starts from point A. From A, turns 60° to reach B (first turn), then to C and so on.

1.    At  third turn, The magnitude of displacement = DG + GA = 500m + 500m = 1000m         Total path length = AB + BC + CD = 500m +500m +500m = 1500 m

2.    At sixth turn, When the motorist takes the 6th turn, he is at starting point A. Therefore,

The magnitude of displacement = 0Total path length = AB + BC + CD + DE + EF + FA = 500m + 500m +500m +500m + 500m + 500m = 3000m

3.     The motorist takes the 8th turn at point C. The magnitude of displacement:

AC= 866.03 m

4.     At sixth turn, When the motorist takes the 6th turn, he is at starting point A. Therefore,

Hence the magnitude of displacement is 866.03m at an angle 30  with AB

Total path length = 600  500 + 500 +500 = 4000 m

4.7

(a) This statement is In order to make a + b+ c + d = 0, it is not necessary to have all four given vectors to be null vectors.

(b) a + b + c +d = 0, ( a + c ) = - (b + d)

Taking the magnitude of both LHS and RHS, | a + c| = | -(b + d)| = | b + d|

So the statement is True.

(c) Given a + b + c + d = 0, a = - (b + c + d)

Taking magnitude of both LHS and RHS, |a| = | -( b + c+ d)| = |b + c +d|

|a| $\le \left|b\right|+\left|c\right|+\left|d\right|$ ……….(1)

Equation (1) shows that the magnitude of a can never be greater than the sum of magnitudes of b,c and d. So the statement is True.

(d) In the given expression a + b + c + d = 0, the sum of 3 vectors a, (b+c), d can be zero only if (b+c) lie in a plane containing a and d, assuming that these three vectors are represented by the three sides of a triangle. If a and d are collinear , then it implies that the vector (b+c) is in the line of a and d. This statement holds good only then the sum of all three vectors will be zero. The statement is True.

(a)

We can write

$\stackrel{?}{\left|OM\right|}$ = $\left|\stackrel{?}{a}\right|\dots \dots \dots .\left(\mathrm{i}\right)$

$$ $\stackrel{?}{\left|MN\right|}$ = $\left|\stackrel{?}{b}\right|$ = $\stackrel{?}{\left|OP\right|}$ ………(ii)

$$ $\stackrel{?}{\left|ON\right|}$ = $\left|\stackrel{?}{a}+\stackrel{?}{b}\right|\dots \dots .\left(iii\right)$

In a triangle, each side is smaller than the sum of the other two sides. Therefore in ΔOMN, we have ON < (OM + MN)

$\left|\stackrel{?}{a}+\stackrel{?}{b}\right|$ < $\left|\stackrel{?}{a}\right|$ + $\left|\stackrel{?}{b}\right|$ ….(iv)

If the two vectors $\stackrel{?}{a}$ and $\stackrel{?}{b}$ act along a straight line in the same direction, then we can write $\left|\stackrel{?}{a}+\stackrel{?}{b}\right|$ = $\left|\stackrel{?}{a}\right|$ + $\left|\stackrel{?}{b}\right|\dots ..\left(v\right)$

Combining equations (iv) and (v), we get

$\left|\stackrel{?}{a}+\stackrel{?}{b}\right|$ $\le$ $\left|\stackrel{?}{a}\right|$ + $\left|\stackrel{?}{b}\right|$

(b)

 

Let two vectors $\stackrel{?}{a}$ and $\stackrel{?}{b}$ be represented by the adjacent sides of a parallelogram OMNP.

We can write

$\stackrel{?}{\left|OM\right|}$ = $\left|\stackrel{?}{a}\right|\dots \dots \dots .\left(\mathrm{i}\right)$

$$ $\stackrel{?}{\left|MN\right|}$ = $\left|\stackrel{?}{b}\right|$ = $\stackrel{?}{\left|OP\right|}$ ………(ii)

$$ $\stackrel{?}{\left|ON\right|}$ = $\left|\stackrel{?}{a}+\stackrel{?}{b}\right|\dots \dots .\left(iii\right)$

In a triangle, each side is smaller than the sum of the other two sides. Therefore in ΔOMN, we have ON + MN > OM and ON + OM > MN

$$ $\stackrel{?}{\left|ON\right|}>\left|\stackrel{?}{OM-OP}\right|$

$\left|\stackrel{?}{a}+\stackrel{?}{b}\right|>$ $\left|\stackrel{?}{a}\right|$ - $\left|\stackrel{?}{b}\right|$ ……(iv)

If the two vectors $\stackrel{?}{a}$ and $\stackrel{?}{b}$ act along a straight line in the same direction, then we can write $\left|\stackrel{?}{a}+\stackrel{?}{b}\right|$ = $\left|\stackrel{?}{a}\right|$ - $\left|\stackrel{?}{b}\right|\dots ..\left(v\right)$

Combining (iv) and (v), we get:

$\left|\stackrel{?}{a}+\stackrel{?}{b}\right|\ge$ $\left|\stackrel{?}{a}\right|$ - $\left|\stackrel{?}{b}\right|$

(c)

Let two vectors $\stackrel{?}{a}$ and $\stackrel{?}{b}$ be represented by the adjacent sides of a parallelogram PORS.

We can write

$$ $\stackrel{?}{\left|OR\right|}$ = $\left|\stackrel{?}{b}\right|$ = $\stackrel{?}{\left|PS\right|}$ ………(i)

$\stackrel{?}{\left|OP\right|}$ = $\left|\stackrel{?}{a}\right|\dots \dots \dots .\left(\mathrm{i}\mathrm{i}\right)$

In a triangle, each side is smaller than the sum of the other two sides. Therefore in ΔOPS, we have OS < OP + PS

$\left|\stackrel{?}{a}-\stackrel{?}{b}\right|$ $<$ $\left|\stackrel{?}{a}\right|$ + $\left|\stackrel{?}{-b}\right|$

$\left|\stackrel{?}{a}-\stackrel{?}{b}\right|$ $<$ $\left|\stackrel{?}{a}\right|$ + $\left|\stackrel{?}{b}\right|$ …….(iii)

If the two vectors $\stackrel{?}{a}$ and $\stackrel{?}{b}$ act along a straight line but in the opposite direction, then we can write $\left|\stackrel{?}{a}-\stackrel{?}{b}\right|$ = $\left|\stackrel{?}{a}\right|$ + $\left|\stackrel{?}{b}\right|\dots ..\left(iv\right)$

Combining (iii) and (iv), we get

$\left|\stackrel{?}{a}-\stackrel{?}{b}\right|$ $\le$ $\left|\stackrel{?}{a}\right|$ + $\left|\stackrel{?}{b}\right|$

(d)

We can write

OS + PS > OP …….(i)

OS > PS – OP …….(ii)

$\left|\stackrel{?}{a}-\stackrel{?}{b}\right|$ $>$ $\left|\stackrel{?}{a}\right|$ + $\left|\stackrel{?}{b}\right|$ ….(iii)

The quantity on the LHS is always positive and that on the RHS can be positive or negative. To make both quantities positive, we take modulus of both sides as:

$\left|\left|\stackrel{?}{a}-\stackrel{?}{b}\right|\right|$ > $\left|\left|\stackrel{?}{a}\right|-\mathrm{}\mathrm{}\left|\stackrel{?}{b}\right|\mathrm{}\right|$ ….(iv)

If the two vectors $\stackrel{?}{a}$ and $\stackrel{?}{b}$ act along a straight line but in the opposite direction, then we can write $\left|\stackrel{?}{a}-\stackrel{?}{b}\right|$ = $\left|\stackrel{?}{a}\right|$ - $\left|\stackrel{?}{b}\right|\dots ..\left(v\right)$

Combining (iv) and (v), we get

$\left|\stackrel{?}{a}-\stackrel{?}{b}\right|$ $\ge$ $\left|\stackrel{?}{a}\right|$ - $\left|\stackrel{?}{b}\right|$