Q: From point A located on a highway (shown in figure above) one has to get by car as soon as possible to point B located in the field at a distance l from the highway. It is known that the car moves in the field \eta times slower than on the highway. At a distance \displaystyle CD=\frac{xl}{\sqrt{\eta^2-1}} from point D one must turn off the highway. Find x .

Let the car turn off the highway at a distance x from the point D . So, CD=x , and if the speed of the car in the field is v , then the time taken by the car to cover the distance AC=AD-x on the highway \displaystyle t_1=\dfrac{AD-x}{\eta v} (1)and the time taken to travel the distance CB in the field \displaystyle t_2=\dfrac{\sqrt{l^2+x^2}}{v} (2)So, the total time elapsed to move the car from point A to B \displaystyle t=t_1+t_2=\dfrac{AD-x}{\eta v}+\dfrac{\sqrt{l^2+x^2}}{v} For t to be minimum \displaystyle\dfrac{dt}{dx}=0 or \displaystyle\dfrac{1}{v}\left[-\dfrac{1}{\eta}+\dfrac{x}{\sqrt{l^2+x^2}}\right]=0 or \eta^2x^2=l^2+x^2 or \displaystyle x=\dfrac{l}{\sqrt{\eta^2-1}} .

Q: Two particles A and B move with velocities v_1\,and\, v_2 respectively along the x and y axis. The initial separation between them is 'd' as shown in the figure. Find the least distance between them during their motion

Let \vec{v_{1}} = v_{1} \hat i \vec{v_{2}} = v_{2} \hat j \vec{v_{2}} - \vec{v_{1}} = v_{2} \hat i - v_{1} \hat j \vec{s_2}(t)-\vec{s_1}(t) = d \hat j + v_2 t \hat i - v_1 t \hat j |\vec{s_2}(t) - \vec{s_1}(t)|^2 = (v_2 t)^2 +(d-v_1t)^2 =(v_2^2+v_1^2)t^2 - 2dv_1t + d^2 = \displaystyle{(\sqrt{v_1^2+v_2^2} t - \dfrac{dv_1}{\sqrt{v_2^2+v_1^2}})^2 + d^2 - \dfrac{d^2v_1^2}{v_1^2+v_2^2} } \geq \dfrac{d^2 v_2^2}{v_1^2+v_2^2} |\vec {s_1} (t) - \vec{s_2}(t)| \geq \dfrac{dv_2}{\sqrt{v_1^2+v_2^2}}

Q: Two particles instantaneously at A and B are 5m , apart and they are moving with uniform velocities, the former towards B at 4m/s and the latter perpendicular to AB at 3m/s . They are nearest at the instant (in seconds)

Velocity of A, { v }_{ A }=4㎧ towards BVelocity of B, { v }_{ B }=3㎧ perpendicular to ABDistance between them, l=5m Let they are closest after time t secondsAnd A and B reach to { A }^{ ' } & { B }^{ ' } respectively.The distance traveled by A =4\times t \therefore { A }^{ ' }B=(5-4t) And distance traveled by B=3\times t \therefore B{ B }^{ ' }=3\times t AAt this distance the distance between A and B is { A }^{ ' }{ B }^{ ' } In \triangle { A }^{ ' }B{ B }^{ ' }:{ \left( { A }^{ ' }{ B }^{ ' } \right) }^{ 2 }={ \left( { A }^{ ' }B \right) }^{ 2 }+{ \left( B{ B }^{ ' } \right) }^{ 2 } ={ (5-4t) }^{ 2 }+{ (3t) }^{ 2 } =25-40t+16{ t }^{ 2 }+9{ t }^{ 2 } =25{ t }^{ 2 }-40t+25 { A }^{ ' }{ B }^{ ' }=\sqrt { { \left( 5t-4 \right) }^{ 2 }+9 } Now { A }^{ ' }{ B }^{ ' } is minimum when { (5t-4) }^{ 2 }=0 \Rightarrow 5t=4 t=\cfrac { 4 }{ 5 } At this time particles are nearest to each other

Q: A particle is thrown vertically upward with a speed of 100 m/s . What is distance of particle travelled in 15 sec ?

\therefore \ t=\dfrac {u}{g}=\dfrac {100}{10}=10\ sec \boxed {t=10\ sec} \therefore \ v=0+10\times 5 \therefore \ v=50\ m/sec Total distance =x+y height, x =\dfrac {u^2}{2g}=\dfrac {100\times 100}{2\times 10}=500\ m y=0(5)+\dfrac {1}{2}\times 10 \times 5^2 y=125\ m \therefore \ total distance =x+y =500+125 =625\ m

Q: A train stops at two stations P and Q which are 2 km apart, It accelerates uniformly from P \ at \ 1 ms^2 for 15 seconds and maintains a constant speed for a time before decelerating uniformly to rest at Q. If the deceleration is 0.5 ms^2 , find the time for which the train is travelling at a constant speed.

SolutionFor first 15s v = u+at [ I^{st} eq^{n} of motion] \Rightarrow v = (0)+2\times 15 v = 30m/s when train is decelerating {v}' = {u}'+{a}'{t}' \Rightarrow (0) = (30)-\dfrac{1}{2}{t}' \Rightarrow {t}' = 60s In first case v^{2}-u^{2} = 2as_{1} [ 3^{rd} eq^{n} of motion] (30)^{2}-(0) = 2(2)s_{1} 225m = s_{1} In third case (0)^{2}-(30)^{2} = 2(-1/2)s_{3} \Rightarrow 900 = s_{3} So, in second case when speed is constant : s_{2} = 2000-s_{1}+s_{2} s_{2} = 2000-900-225 s_{2} = 875m s_{2} = v{t}'' {t}'' = \dfrac{s_{2}}{v} = \dfrac{875}{30} = 29.16s

Q: Starting from rest a particle moves in a straight line with acceleration a = (25 - t^2)^{1/2} m/s^2 \ for \ 0 \leq t \leq 5s a = \dfrac{3 \pi}{8} m/s^2 \ for \ t \geq 5s The velocity of particle at t = 7s is:

v=\int a\, dt a = (25 - t^2)^{1/2} m/s^2 \ for \ 0 \leq t \leq 5s = \dfrac{3 \pi}{8} m/s^2 \ for \ t \geq 5s v =\int(25 - t^2)^{1/2} \,dt+a\ \ \ \ \ \ \ \ 0 \leq t \leq 5s = \int\dfrac{3 \pi}{8} \,dt +b\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t \geq 5s v =\frac{t}{2}\sqrt{5^2-t^2}+\dfrac{25}{2}sin^{-1}(\dfrac{t}{5})+a\ \ \ \ \ \ \ \ 0 \leq t \leq 5s = \dfrac{3 \pi t}{8} +b\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t \geq 5s at t=0 , v=0 gives a=0 finding b using the fact that speed is a continous fuction lim_{t\to 5^{-}} v=lim_{t\to 5^+}v 0+\dfrac{25}{2}sin^{-1}(\dfrac{5}{5})=\dfrac{3\pi \times 5}{8}+b b=\dfrac{35\pi}{8} v =\frac{t}{2}\sqrt{5^2-t^2}+\dfrac{25}{2}sin^{-1}(\dfrac{t}{5})\ \ \ \ \ \ \ \ 0 \leq t \leq 5s = \dfrac{3 \pi t}{8} +\dfrac{35\pi}{8}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t \geq 5s so at t=7 v= \dfrac{3 \pi \times 7}{8} +\dfrac{35\pi}{8}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v= \dfrac{8 \pi \times 7}{8} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v=7\pi v=22m/s

Q: A street car moves rectilinearly from station A (here car stops) to next station B (here also car stops) with an acceleration varying according to the law f = a - bx , where a and b are positive constants and x is the distance from station A . If the maximum distance between the two stations is x = \dfrac {Na}{b} then find N .

Given that, The acceleration is f=a-bx..... (I) The acceleration is decreasing with increasing x. Hence, the velocity will be maximum Where, f= 0, From equation (I), a-bx=0 x=\dfrac{a}{b} Now, we know that f=\dfrac{dv}{dt} f=\dfrac{dv}{dt}\times \dfrac{dx}{dx} f=v\dfrac{dv}{dx} Now, we can write, v\dfrac{dv}{dx}=a-bx vdv=(a-bx)dx.....(II) Now, for maximum velocity, \int\limits_{0}^{{{v}_{\max }}}{vdv}=\int\limits_{0}^{\dfrac{a}{b}}{(a-bx)dx} \left[ \dfrac{{{v}^{2}}}{2} \right]_{0}^{{{v}_{\max }}}=\left( ax-b\dfrac{{{x}^{2}}}{2} \right)_{0}^{\dfrac{a}{b}} \dfrac{v_{\max }^{2}}{2}=\dfrac{{{a}^{2}}}{b}-\dfrac{b{{a}^{2}}}{2{{b}^{2}}} \dfrac{v_{\max }^{2}}{2}=\dfrac{2b{{a}^{2}}-b{{a}^{2}}}{2{{b}^{2}}} \dfrac{v_{\max }^{2}}{2}=\dfrac{{{a}^{2}}}{2b} v_{\max }^{2}=\dfrac{{{a}^{2}}}{b} {{v}_{\max }}=\dfrac{a}{\sqrt{b}} Now, for the maximum distances x between the two stations On integrating of equation (II) At A and B, cars at rest so the velocity is zero at the both stations. Now, \int\limits_{0}^{0}{vdv}=\int\limits_{0}^{x}{\left( a-bx \right)dx} 0=\left( ax-b\dfrac{{{x}^{2}}}{2} \right)_{0}^{x} ax-b\dfrac{{{x}^{2}}}{2}=0 a-b\dfrac{x}{2}=0 x=\dfrac{2a}{b}......(III) And given that, x=\dfrac{Na}{b}....(IV) Now, on comparing equation (III) and (IV) Then, N=2 Hence, the value of N is 2 .

Q: A hunter is at (4,-1,5) units. He observes two preys at P_{1}(-1,2,0) units and P_{2}(1,1,4) respectively. At zero instant he starts moving in the plane of their positions with uniform speed of 5\: units \: s^{-1} in a direction perpendicular to line P_{1}P_{2} till he sees P_{1} and P_{2} collinear at time T . Time T is

Let,A be the vector acting along hunter and P_1 andB be the vector acting along P_1 and P_2 Also,R be the vector of velocity with which hunter travels perpendicular to P_1.P_2 Then, \vec {A}=(-1-4) \hat{i}+(2+1) \hat{j}+(0-5)\hat{k}=-5\hat{i}+3 \hat{j}- 5 \hat{k} and \vec {B}=(1+1) \hat{i}+(1-2) \hat{j}+(4-0)\hat{k}=2 \hat{i}- \hat{j}+4 \hat{k} Now, R=|\vec {A}| \sin \theta and |\vec {A}\times\vec {B}|=AB \sin \theta or \sin \theta=\displaystyle \frac{|\vec{A} \times \vec{B}|}{AB} i.e R=A \displaystyle \frac{|\vec{A} \times \vec{B}|}{AB}=\displaystyle \frac{|\vec{A} \times \vec{B}|}{B} Now, \vec {A} \times \vec {B}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ -5 & 3 & -5\\ 2 & -1 & 4\end{vmatrix} =(12-5)\hat{i} + (-10+20)\hat{j} + (5-6) \hat{k}=7\hat{i}+10\hat{j}-\hat{k} |\vec {A} \times \vec {B}|=(7^{2}+10^{2}+1)^{1/2}=12.25 |\vec {B}|=(2^{2}+1^{2}+4^{2})^{1/2}=4.58 \therefore R=\displaystyle \frac{12.25}{4.58}=2.67 \:m Time taken to reach destination, =\displaystyle \frac{R}{v}=\displaystyle \frac{2.67}{5}=0.53 \:s

Question 1

The position of a particle travelling along x axis is given by  x_{1} = t^{3}- 9t^{2} + 6t  where  x_{t}  is in  cm  and  t  is in second. Then

Accepted Answer

x_{t}=t^{3}-9t^{2}+6t V=\displaystyle \frac{d_{X_{t}}}{dt}=(3t^{2}-18t+6)cms^{-1} For body to be at rest  v = 0 \Rightarrow t=(3\pm\sqrt{7})s \Rightarrow t_{1}=(3-\sqrt{7})s  and  \Rightarrow t_{2}=(3+\sqrt{7})s Displacement of particle between  t_1  and  t_2  is  {x_t}_2 - {x_t}_1 \displaystyle ({(3+\sqrt{7})}^3 -9{(3+\sqrt{7})}^2 +6{(3+\sqrt{7})}) - ({(3-\sqrt{7})}^3 - 9{(3-\sqrt{7})}^2 + 6{(3-\sqrt{7})}) = -74 v< 0\, for\, (3-\sqrt{7})< t< (3+\sqrt{7})

Question 2

From point  A  located on a highway (shown in figure above) one has to get by car as soon as possible to point  B  located in the field at a distance  l  from the highway. It is known that the car moves in the field  \eta  times slower than on the highway. At a distance  \displaystyle CD=\frac{xl}{\sqrt{\eta^2-1}}  from point  D  one must turn off the highway. Find  x .

Accepted Answer

Let the car turn off the highway at a distance  x  from the point  D . So,  CD=x , and if the speed of the car in the field is  v , then the time taken by the car to cover the distance  AC=AD-x  on the highway \displaystyle t_1=\dfrac{AD-x}{\eta v}  (1)and the time taken to travel the distance  CB  in the field \displaystyle t_2=\dfrac{\sqrt{l^2+x^2}}{v}  (2)So, the total time elapsed to move the car from point  A  to  B \displaystyle t=t_1+t_2=\dfrac{AD-x}{\eta v}+\dfrac{\sqrt{l^2+x^2}}{v} For  t  to be minimum \displaystyle\dfrac{dt}{dx}=0  or  \displaystyle\dfrac{1}{v}\left[-\dfrac{1}{\eta}+\dfrac{x}{\sqrt{l^2+x^2}}\right]=0 or  \eta^2x^2=l^2+x^2  or  \displaystyle x=\dfrac{l}{\sqrt{\eta^2-1}} .

Question 3

Question 4

Two particles A and B move with velocities  v_1\,and\, v_2   respectively along the x and y axis.  The initial separation between them is 'd' as shown in the figure. Find the least distance between them during their motion

Accepted Answer

Let  \vec{v_{1}} = v_{1} \hat i \vec{v_{2}} = v_{2} \hat j \vec{v_{2}} - \vec{v_{1}} = v_{2} \hat i - v_{1} \hat j \vec{s_2}(t)-\vec{s_1}(t) = d \hat j + v_2 t \hat i - v_1 t \hat j |\vec{s_2}(t) - \vec{s_1}(t)|^2 = (v_2 t)^2 +(d-v_1t)^2                               =(v_2^2+v_1^2)t^2 - 2dv_1t + d^2                               = \displaystyle{(\sqrt{v_1^2+v_2^2}  t - \dfrac{dv_1}{\sqrt{v_2^2+v_1^2}})^2 + d^2 - \dfrac{d^2v_1^2}{v_1^2+v_2^2} }                               \geq \dfrac{d^2 v_2^2}{v_1^2+v_2^2} |\vec {s_1} (t) - \vec{s_2}(t)| \geq \dfrac{dv_2}{\sqrt{v_1^2+v_2^2}}

Question 5

Two particles instantaneously at  A  and  B  are  5m , apart and they are moving with uniform velocities, the former towards  B  at  4m/s  and the latter perpendicular to  AB  at  3m/s . They are nearest at the instant (in seconds)

Accepted Answer

Velocity of A,  { v }_{ A }=4&#13223;  towards BVelocity of B,  { v }_{ B }=3&#13223;  perpendicular to ABDistance between them,  l=5m Let they are closest after time t secondsAnd A and B reach to  { A }^{ ' }  &  { B }^{ ' } respectively.The distance traveled by A =4	imes t 	herefore { A }^{ ' }B=(5-4t) And distance traveled by  B=3	imes t 	herefore B{ B }^{ ' }=3	imes t AAt this distance the distance between A and B is  { A }^{ ' }{ B }^{ ' } In  	riangle { A }^{ ' }B{ B }^{ ' }:{ \left( { A }^{ ' }{ B }^{ ' } ight)  }^{ 2 }={ \left( { A }^{ ' }B ight)  }^{ 2 }+{ \left( B{ B }^{ ' } ight)  }^{ 2 } ={ (5-4t) }^{ 2 }+{ (3t) }^{ 2 } =25-40t+16{ t }^{ 2 }+9{ t }^{ 2 } =25{ t }^{ 2 }-40t+25 { A }^{ ' }{ B }^{ ' }=\sqrt { { \left( 5t-4 ight)  }^{ 2 }+9 }  Now  { A }^{ ' }{ B }^{ ' }  is minimum when  { (5t-4) }^{ 2 }=0 \Rightarrow 5t=4 t=\cfrac { 4 }{ 5 }  At this time particles are nearest to each other

Question 6

A particle is thrown vertically upward with a speed of 100 m/s . What is distance of particle travelled in 15 sec ?

Accepted Answer

herefore \ t=\dfrac {u}{g}=\dfrac {100}{10}=10\ sec \boxed {t=10\ sec} 	herefore \ v=0+10	imes 5 	herefore \ v=50\ m/sec Total distance  =x+y height, x   =\dfrac {u^2}{2g}=\dfrac {100	imes 100}{2	imes 10}=500\ m y=0(5)+\dfrac {1}{2}	imes 10 	imes 5^2 y=125\ m 	herefore \   total distance  =x+y =500+125 =625\ m

Question 7

A train stops at two stations P and Q which are 2 km apart, It accelerates uniformly from   P \  at \ 1 ms^2   for 15 seconds and maintains a constant speed for a time before decelerating uniformly to rest at Q. If the deceleration is   0.5 ms^2  , find the time for which the train is travelling at a constant speed.

Accepted Answer

SolutionFor first 15s  v = u+at          [  I^{st} eq^{n}  of motion]  \Rightarrow v = (0)+2	imes 15   v = 30m/s  when train is decelerating   {v}' = {u}'+{a}'{t}'   \Rightarrow (0) = (30)-\dfrac{1}{2}{t}'   \Rightarrow {t}' = 60s  In first case   v^{2}-u^{2} = 2as_{1}     [  3^{rd} eq^{n}   of motion]  (30)^{2}-(0) = 2(2)s_{1}   225m = s_{1}  In third case  (0)^{2}-(30)^{2} = 2(-1/2)s_{3}   \Rightarrow 900 = s_{3}  So, in second case when speed is constant :  s_{2} = 2000-s_{1}+s_{2}   s_{2} = 2000-900-225   s_{2} = 875m   s_{2} = v{t}''  {t}'' = \dfrac{s_{2}}{v} = \dfrac{875}{30} = 29.16s

Question 8

Starting from rest a particle moves in a straight line with acceleration

a = (25 - t^{2})^{1 / 2} m / s^{2} f o r 0 \leq t \leq 5 s

a = \frac{3 π}{8} m / s^{2} f o r t \geq 5 s

The velocity of particle at t = 7s is:

Accepted Answer

v=\int a\, dt a = (25 - t^2)^{1/2} m/s^2 \ for \ 0 \leq t \leq 5s      = \dfrac{3 \pi}{8} m/s^2 \ for \ t \geq 5s v =\int(25 - t^2)^{1/2} \,dt+a\ \  \  \  \ \ \ \ 0 \leq t \leq 5s      = \int\dfrac{3 \pi}{8} \,dt +b\ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \  \ \ \ \ \ t \geq 5s v =\frac{t}{2}\sqrt{5^2-t^2}+\dfrac{25}{2}sin^{-1}(\dfrac{t}{5})+a\ \  \  \  \ \ \ \ 0 \leq t \leq 5s      = \dfrac{3 \pi t}{8}  +b\ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \  \ \ \ \ \ t \geq 5s at  t=0 , v=0  gives  a=0 finding  b  using the fact that speed is a continous fuction lim_{t	o 5^{-}}  v=lim_{t	o 5^+}v 0+\dfrac{25}{2}sin^{-1}(\dfrac{5}{5})=\dfrac{3\pi 	imes 5}{8}+b b=\dfrac{35\pi}{8} v =\frac{t}{2}\sqrt{5^2-t^2}+\dfrac{25}{2}sin^{-1}(\dfrac{t}{5})\ \  \  \  \ \ \ \ 0 \leq t \leq 5s      = \dfrac{3 \pi t}{8}  +\dfrac{35\pi}{8}\ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \  \ \ \ \ \ t \geq 5s so at  t=7    v= \dfrac{3 \pi 	imes 7}{8}  +\dfrac{35\pi}{8}\ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \  \ \ \ \ \     v= \dfrac{8 \pi 	imes 7}{8} \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \  \ \ \ \ \  v=7\pi v=22m/s

Question 9

A street car moves rectilinearly from station  A  (here car stops) to next station  B  (here also car stops) with an acceleration varying according to the law  f = a - bx , where  a  and  b  are positive constants and  x  is the distance from station  A . If the maximum distance between the two stations is  x = \dfrac {Na}{b}  then find  N .

Accepted Answer

Given that,&#10;&#10;The acceleration is  f=a-bx..... (I) &#10;&#10;The acceleration is decreasing with increasing x.&#10;&#10;Hence, the velocity will be maximum&#10;&#10;Where, f= 0,&#10;&#10;From equation (I),&#10;&#10;    a-bx=0   &#10;&#10;   x=\dfrac{a}{b}   &#10;&#10;Now, we know that&#10;&#10;    f=\dfrac{dv}{dt}  &#10;&#10;&#10;   f=\dfrac{dv}{dt}	imes&#10;\dfrac{dx}{dx}   &#10;&#10;   f=v\dfrac{dv}{dx}  &#10;&#10;&#10;Now, we can write,&#10;&#10;    v\dfrac{dv}{dx}=a-bx&#10;  &#10;&#10;  &#10;vdv=(a-bx)dx.....(II)   &#10;&#10;Now, for maximum velocity,&#10;&#10;   &#10;\int\limits_{0}^{{{v}_{\max }}}{vdv}=\int\limits_{0}^{\dfrac{a}{b}}{(a-bx)dx}  &#10;&#10;&#10;   \left[ \dfrac{{{v}^{2}}}{2}&#10;ight]_{0}^{{{v}_{\max }}}=\left( ax-b\dfrac{{{x}^{2}}}{2} ight)_{0}^{\dfrac{a}{b}}&#10;  &#10;&#10;   \dfrac{v_{\max&#10;}^{2}}{2}=\dfrac{{{a}^{2}}}{b}-\dfrac{b{{a}^{2}}}{2{{b}^{2}}}   &#10;&#10;   \dfrac{v_{\max&#10;}^{2}}{2}=\dfrac{2b{{a}^{2}}-b{{a}^{2}}}{2{{b}^{2}}}   &#10;&#10;   \dfrac{v_{\max&#10;}^{2}}{2}=\dfrac{{{a}^{2}}}{2b}   &#10;&#10;   v_{\max }^{2}=\dfrac{{{a}^{2}}}{b}&#10;  &#10;&#10;   {{v}_{\max }}=\dfrac{a}{\sqrt{b}}&#10;  &#10;&#10;Now, for the maximum distances x between the two stations&#10;&#10;On integrating of equation (II)&#10;&#10;At A and B, cars at rest so the velocity is zero at the&#10;both stations.&#10;&#10;Now,&#10;&#10;   &#10;\int\limits_{0}^{0}{vdv}=\int\limits_{0}^{x}{\left( a-bx ight)dx}   &#10;&#10;   0=\left( ax-b\dfrac{{{x}^{2}}}{2}&#10;ight)_{0}^{x}   &#10;&#10;   ax-b\dfrac{{{x}^{2}}}{2}=0&#10;  &#10;&#10;   a-b\dfrac{x}{2}=0  &#10;&#10;&#10;   x=\dfrac{2a}{b}......(III)&#10;  &#10;&#10;And given that,&#10;&#10; x=\dfrac{Na}{b}....(IV) &#10;&#10;Now, on comparing equation (III) and (IV)&#10;&#10;Then,&#10;&#10; N=2 Hence, the value of  N  is  2 .&#10;&#10;

Question 10

A hunter is at  (4,-1,5)  units. He observes two preys at  P_{1}(-1,2,0)  units and  P_{2}(1,1,4)  respectively. At zero instant he starts moving in the plane of their positions with uniform speed of  5\: units \: s^{-1}  in a direction perpendicular to line  P_{1}P_{2}  till he sees  P_{1}  and  P_{2}  collinear at time  T . Time  T  is

Accepted Answer

Let,A be the vector acting along hunter and  P_1  andB be the vector acting along  P_1  and  P_2 Also,R be the vector of velocity with which hunter travels perpendicular to  P_1.P_2 Then, \vec {A}=(-1-4) \hat{i}+(2+1) \hat{j}+(0-5)\hat{k}=-5\hat{i}+3 \hat{j}- 5 \hat{k}  and \vec {B}=(1+1) \hat{i}+(1-2) \hat{j}+(4-0)\hat{k}=2 \hat{i}- \hat{j}+4 \hat{k}  Now,  R=|\vec {A}| \sin 	heta  and  |\vec {A}	imes\vec {B}|=AB \sin 	heta  or  \sin 	heta=\displaystyle \frac{|\vec{A} 	imes \vec{B}|}{AB} i.e  R=A \displaystyle \frac{|\vec{A} 	imes \vec{B}|}{AB}=\displaystyle \frac{|\vec{A} 	imes \vec{B}|}{B} Now,  \vec {A} 	imes \vec {B}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \  -5 & 3 & -5\ 2 & -1 & 4\end{vmatrix} =(12-5)\hat{i} + (-10+20)\hat{j} + (5-6) \hat{k}=7\hat{i}+10\hat{j}-\hat{k} |\vec {A} 	imes \vec {B}|=(7^{2}+10^{2}+1)^{1/2}=12.25 |\vec {B}|=(2^{2}+1^{2}+4^{2})^{1/2}=4.58 	herefore R=\displaystyle \frac{12.25}{4.58}=2.67 \:m Time taken to reach destination, =\displaystyle \frac{R}{v}=\displaystyle \frac{2.67}{5}=0.53 \:s

Motion In A Straight Line