Q: In a harbour, wind is blowing at the speed of 72km.h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. It the boat starts moving at a speed of 51km/h to the north, what is the direction of the flag on the mast of the boat?

The angle between v_w and (-v_b)=90+45 \tan{\beta}=\dfrac{51\sin{90+45}}{72+51\cos{90+45}} Substituting and solving we get- \tan{\beta}=\dfrac{51}{50.80}=1.0038 \beta=\tan^{-1}{1.0038}=45.11^{\circ} Angle with respect to the east direction =45.11^{\circ}-45^{\circ}=0.11^{\circ} Hence the flag will flutter almost due east.

Q: A body is moving along a straight line. Its distance { X }_{ t } from a point on its path at a time t after passing that point is given by { X }_{ t }=8{ t }^{ 2 }-3{ t }^{ 3 } , where { X }_{ t } is in meter and t is in second. The correct statement(s) is/are:

\Rightarrow { X }_{ t }=8{ t }^{ 2 }-3{ t }^{ 3 }\\ \Rightarrow \dfrac { d{ X }_{ t } }{ dt } =V=16t-9{ t }^{ 2 }\\ \Rightarrow \dfrac { { d }^{ 2 }{ X }_{ t } }{ { dt }^{ 2 } } =a=16-18t Now, { X }_{ t }=8{ t }^{ 2 }-3{ t }^{ 3 }\\ At\ t=4\ seconds,\\ { X }_{ t }=-64\ meters\\ Average\ velocity=\dfrac { -64 }{ 4 } =-16 Now, \Rightarrow \dfrac { d{ X }_{ t } }{ dt } =V=16t-9{ t }^{ 2 }\\ \Rightarrow V=t(16-9t)\\ \Rightarrow V=0\quad at\ t=0 and\ t=\dfrac { 16 }{ 9 } seconds.\\ \Rightarrow The\ object\ changes\ its\ velocity\ in\ opposite\ direction\ at\ t=\frac { 16 }{ 9 } seconds. Now, \Rightarrow \dfrac { { d }^{ 2 }{ X }_{ t } }{ { dt }^{ 2 } } =a=16-18t\\ At\ t=4\ seconds,\\ \Rightarrow a=16-(18\times 4)=-56 Now, Total\ Distance\ travelled\ by\ object:\\ between\ t=0\ to\ t=\dfrac { 16 }{ 9 } seconds\Rightarrow d=8.42\\ \therefore Total\ distance\ travelled\ between\ t=0\ to\ t=4\ seconds\ is\\ \Rightarrow 64+(2\times 8.42)=80.84\ meters.\\ \Rightarrow Average\ speed=\dfrac { 80.84 }{ 4 } =20.21\quad

Question 1

A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes.

Accepted Answer

Given,Total distance covered  10m  in  40sec So,Total time taken  = 2min c= 120sec Total round completed = \dfrac{120}{40}=3 rounds So If an object moves from A to D, the shortest path will be from A to C. So displacement is AC, that we will find by phythogoras theorem. AC^2= AB^2 +BC^2   AC^2= (10)^2 +(10)^2 AC^2= 100+100 AC^2=200 AC=\sqrt{200}=10\sqrt2m

Question 2

In a harbour, wind is blowing at the speed of  72km.h  and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. It the boat starts moving at a speed of  51km/h  to the north, what is the direction of the flag on the mast of the boat?

Accepted Answer

The angle between  v_w and (-v_b)=90+45 	an{\beta}=\dfrac{51\sin{90+45}}{72+51\cos{90+45}} Substituting and solving we get- 	an{\beta}=\dfrac{51}{50.80}=1.0038 \beta=	an^{-1}{1.0038}=45.11^{\circ} Angle with respect to the east direction  =45.11^{\circ}-45^{\circ}=0.11^{\circ} Hence the flag will flutter almost due east.

Question 3

A body is moving along a straight line. Its distance

X_{t}

from a point on its path at a time

t

after passing that point is given by

X_{t} = 8 t^{2} - 3 t^{3}

, where

X_{t}

is in meter and

t

is in second. The correct statement(s) is/are:

Accepted Answer

\Rightarrow  { X }_{ t }=8{ t }^{ 2 }-3{ t }^{ 3 }\ \Rightarrow  \dfrac { d{ X }_{ t } }{ dt } =V=16t-9{ t }^{ 2 }\ \Rightarrow  \dfrac { { d }^{ 2 }{ X }_{ t } }{ { dt }^{ 2 } } =a=16-18t Now, { X }_{ t }=8{ t }^{ 2 }-3{ t }^{ 3 }\ At\ t=4\ seconds,\ { X }_{ t }=-64\ meters\ Average\ velocity=\dfrac { -64 }{ 4 } =-16  Now, \Rightarrow  \dfrac { d{ X }_{ t } }{ dt } =V=16t-9{ t }^{ 2 }\ \Rightarrow  V=t(16-9t)\ \Rightarrow  V=0\quad at\ t=0 and\ t=\dfrac { 16 }{ 9 } seconds.\ \Rightarrow  The\ object\ changes\ its\ velocity\ in\ opposite\ direction\ at\ t=\frac { 16 }{ 9 } seconds. Now, \Rightarrow  \dfrac { { d }^{ 2 }{ X }_{ t } }{ { dt }^{ 2 } } =a=16-18t\ At\ t=4\ seconds,\ \Rightarrow a=16-(18	imes 4)=-56  Now, Total\ Distance\ travelled\ by\ object:\ between\ t=0\ to\ t=\dfrac { 16 }{ 9 } seconds\Rightarrow  d=8.42\ 	herefore  Total\ distance\ travelled\ between\ t=0\ to\ t=4\ seconds\ is\ \Rightarrow  64+(2	imes 8.42)=80.84\ meters.\ \Rightarrow  Average\ speed=\dfrac { 80.84 }{ 4 } =20.21\quad

Question 4

Particles A and B are moving with constant velocities along  x  and  y  axis respectively, the graph of separation between them with time is

Accepted Answer

their distance will be given by  \sqrt{x^2+y^2} x_0=-x  and  y_0=-y  and velocities  v_x=v_y=C .then  x_t=-x+C.t  and  y_t=-y+C.t distance is given by  D^2=(-x+C.t)^2+(-y+C.t)^2 this is quadratic between  D^2  and  t  so best possible curve is option  D .

Question 5

The velocity-time graph of a body is shown in figure.

Accepted Answer

From  t=0  to  t=A  velocity increases from 0 to v where v is the velocity at  t=A From  t=A  to  t=B  velocity increases from v to 0 where v is the velocity at  t=A Hence,  v_{avg}  during interval  0 to A is same as  v_{avg}  during interval  A to B  \frac{v_c-0}{A-0} =	an 60^{\circ}=\sqrt{3} ;  	herefore v_c = \sqrt{3} 	imes OA \frac{v_c-0}{A-0} =	an 150^{\circ}=-	an 30^{\circ}=-\frac{1}{\sqrt{3}}  	herefore v_c = \frac{1}{\sqrt {3}} 	imes AB Equating both  v_c ,  \sqrt{3} 	imes OA = \frac{1}{\sqrt{3}} 	imes AB  \Rightarrow \frac{AB}{OA} =3 \Rightarrow  \frac{OA}{OA + AB} =\frac{1}{1+3}  \Rightarrow \frac{OA}{OB} =\frac{1}{4} Distance covered on in time OA:  d_{OA} = \frac{1}{2} 	imes OA 	imes v_A Distance covered on in time AB:  d_{AB} = \frac{1}{2} 	imes AB 	imes v_A |a_{OA} |=	an 60^{\circ}=\sqrt{3} |a_{AB}| =	an 30^{\circ}\frac{1}{\sqrt{3}} \frac{|a_{OA} |}{|a_{AB}| } =3  	herefore \frac{d_{OA}}{d_{AB}} = \frac{\frac{1}{2} 	imes OA 	imes v_A}{\frac{1}{2} 	imes AB 	imes v_A} 	herefore \frac{d_{OA}}{d_{AB}}=\frac{ \frac{AC}{\cot 60^{\circ}}}{\frac{AC}{\cot 30^{\circ}}}=	an 60^{\circ} 	imes \cot 30^{\circ}=3

Question 6

Two stones are thrown vertically upwards simultaneously from the same point on the ground with initial speed

u_{1} = 30

m/sec and

u_{2} = 50 m / s e c .

Which of the curves represent correct variation (for the time interval in which both reach the ground) of relative position

(x_{2} - x_{1})

and relative velocity

(v_{2} - v_{1})

of second stone with respect to respect to first with time (t). Assume that stones do not rebound after hitting.

Accepted Answer

While both the stones are in flight,  a_{1}=g\ and\ a_{2}=g So  a_{rel}=0 \Rightarrow V_{rel}= constant \Rightarrow X_{rel}= \left ( const \right )t \Rightarrow Curve\ of x_{rel}v/s   t will be straight line.After the first particle drops on ground, the seperation  \left ( x_{rel} \right )  will decrease parabolically (due to gravitational acceleration),and finally becomes zero.and  V_{rel}= slope\ of\ x_{rel} v/s t

Question 7

A body starts from rest and moves with an uniform acceleration. The ratio of distance covered in the  n^{th}  second to the distance covered in  n  seconds is :

Accepted Answer

Distance covered in  n  seconds,o  \rightarrow  start from rest s = ut + \dfrac {1}{2} at^2 s = \dfrac {1}{2} an^2 -----(1) Distance covered in  n^{th}  second  = s_n - s_{n-1} \Rightarrow s_n - s_{n-1} = \dfrac {1}{2} a n^2 - \dfrac {1}{2} a (n - 1)^2 ----- (2) So,  \dfrac {eq^n (2)}{eq^n (1)} = \dfrac {\dfrac {1}{2}an^2 - \dfrac {1}{2} a(n-1)^2}{\dfrac {1}{2}an^2} = \dfrac {2n - 1}{n^2} = \dfrac {2}{n} - \dfrac {1}{n^2}

Question 8

The acceleration of a car started at  t = 0 , varies with time as shown in figure the distance travelled by the car in the  25 \ s  is:

Accepted Answer

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Motion In A Straight Line