Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.
Problem: 1.13
Point \(A\) moves uniformly with velocity \(v\) so that the vector \(v\) is continually "aimed" at point \(B\) which in its turn moves rectilinearly and uniformly with velocity \(u\). At the initial moment of time \(v\) perpendicular to \(u\) and the points are separated by a distance \(L\). How soon will the point converge?
Solution: 1.13
Figure shows an typical instantaneous state of the two particles during their motion as particle \(B\) is continuously aimed towards particle \(A\). Let \(x\) be the horizontal distance between the two particles and \(y\) be the vertical distance between the two particles. From second figure we can write the motion of the particle \(B\) given by the following equations,
\(\frac{{dx}}{{dt}} = u - v\cos \theta \) ... ... ... (i)
\(\frac{{dy}}{{dt}} = - v\sin \theta \) ... ... ... (ii)
Now, integrating both side (i) from \(t = 0\) to the time when \(A\) and \(B\) meet we obtain,
\(\int_0^0 {dx} = \int_0^\tau {v\cos \theta } - u\tau \)
Or, \(0 = \int_0^\tau {v\cos \theta } - u\tau \)
Or, \(\int_0^\tau {\cos \theta dt} = \frac{{u\tau }}{v}\) ... ... ... (iii)
The integral is from \(0\) to \(0\) in the above equation because the horizontal distance between \(A\) and \(B\) initially is \(0\) and finally when they meet also it will again \(0\). This can seen from the figure, from the definition of \(x\) as the horizontal distance between \(A\) and \(B\).
In the figure, we resolve all velocity along and perpendicular to the line connecting the two bodies.
If \(r\) is the distance between the two bodies then, as seen from the figure \(r\) decreases as \(\left( {v - u\cos \theta } \right)\) at any given instant of time.
Thus,
\(\frac{{dr}}{{dt}} = - \left( {v - u\cos \theta } \right)\)
The same result can of course also be derived by transforming to polar co-ordinates. Now if \(r\) is the vector connecting particles \(A\) and \(B\) then,
\(x = r\cos \theta \) and \(y = r\sin \theta \)
Or, \(\frac{{dx}}{{dt}} = \frac{{dr}}{{dt}}\cos \theta - r\sin \theta \frac{{d\theta }}{{dt}}\)
And, \(\frac{{dy}}{{dt}} = \frac{{dr}}{{dt}}\sin \theta + r\cos \theta \frac{{d\theta }}{{dt}}\)
Therefore,
\(\frac{{dr}}{{dt}}\cos \theta - r\sin \frac{{d\theta }}{{dt}} = u - v\cos \theta \)
And, \(\frac{{dr}}{{dt}}\sin \theta + r\cos \theta = - v\sin \theta \)
Or, \(\frac{{dr}}{{dt}}{\cos ^2}\theta - r\sin \theta \cos \theta = u\cos \theta - v{\cos ^2}\theta \)
And, \(\frac{{dr}}{{dt}}{\sin ^2}\theta + r\cos \theta \sin \theta \frac{{d\theta }}{{dt}} = - v{\sin ^2}\theta \)
Therefore,
\(\frac{{dr}}{{dt}} = u\cos \theta - v\)
Integrating both side, it would mean that,
\(\int_L^0 {dr} = \int_0^\tau {\left( {u\cos \theta - v} \right)dt} \)
Or, \(L = \int_0^\tau {\left( {v - u\cos \theta } \right)dt} \)
Or, \(L = v\tau - u\int_0^\tau {\cos \theta dt} \)
Now we can use the equation (ii) to solve for \(\tau \) as,
\(L = v\tau - u\left( {\frac{{u\tau }}{v}} \right)\)
Or, \(\tau = \frac{{Lv}}{{\left( {{v^2} - {u^2}} \right)}}\)
Problem: 1.12
Three points are located at the vertices of an equilateral triangle whose side equals \(a\). They all atart moving simultaneously with velocity \(v\) constant in modulus, with the first point heading continually for the second, the second for the third, and the third for the first. How soon will the points converge?
Solution: 1.12
Thus, the motion of particles will depict a shrinking equilateral triangle that is also rotating about the centroid as it shrinks. This is depicted in the figure.
To an observer located on one of the particles the components of velocities of its adjacent particle at any given point of time are depicted in the third figure. Thus, any given particles feel that while it is moving towards its adjacent particle at a speed \(v\), the adjacent particle is moving towards it at a speed \(\frac{v}{2}\). Thus the particles seem to approach each other at a constant rate of \(\left( {v + \frac{v}{2}} \right) = \frac{{3v}}{2}\). Since the initial distance was \(a\), the time takes to meet is given by \(t = \frac{a}{{\frac{{3v}}{2}}} = \frac{{2a}}{{3v}}\).
Problem: 1.11
Two particles move in a uniform gravitational field with an acceleration \(g\). At the initial moment the particles were located at one point and moved with velocities \({v_1} = 3m/s\) and \({v_2} = 4m/s\) horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors becomes mutually perpendicular.
Solution: 1.11
Here the two particle that was shot with \({v_1} = 3m/s\) and the one that was shot with \({v_2} = 4m/s\). Since \({v_1}\) and \({v_2}\) are supposed to be in opposite directions, we choose \({v_1}\) as the positive directions and \({v_2}\) as the negative direction. Further, we imagine that gravity acting downward is the negative direction.
Since, gravity acts vertically downwards, it does not effect the horizontal components of velocities. However both the particle accelerate at equal rates of \(g\) downwards from zero initial velocity component in downward direction.
The velocity of first particle after time \(t\) is given by \(\left( {{v_1}\hat i - gt\hat j} \right)\) and that second particle is given by \(\left( {{-
v_2}\hat i - gt\hat j} \right)\).
When the two velocity vectors are perpendicular to each other, the dot product of the vectors will be zero.
That is,
\(\left( {{v_1}\hat i - gt\hat j} \right).\left( { - {v_2}\hat i - gt\hat j} \right) = 0\)
Or, \(\left( { - {v_1}{v_2} + {g^2}{t^2}} \right) = 0\)
Or, \(\left( { - 12 + 100{t^2}} \right) = 0\)
Or, \(t = \sqrt {0.12} \)
The vertical distance traveled by the two particle will be identical, thus at any time \(t\) the distance between the two particles is only a function of the horizontal components of their positions is given by \(\left( {{v_1} + {v_2}} \right)t\).
Thus the distance between the two particles when their velocity vectors are mutually perpendicular are given by,
\( = \left( {{v_1} + {v_2}} \right) \times t = 7 \times \sqrt {0.12} = 2.5m\)
Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.
Problem: 1.10
A boat moves relative to water with a velocity which is \(n = 2.0\) times less than the river flow velocity. At what angle to the stream direction must the boat move to minimize drifting?
Solution: 1.10
The figure shows the components of the velocities of the two bodies along the horizontal and vertical components. The position of the vertically thrown body after time \(t\) will be \(0\hat i + \left( {{v_0}t - \frac{1}{2}g{t^2}} \right)\hat j\), where \(\hat i\) and \(\hat j\) are the unit vectors along the horizontal and vertical directions respectively.
The position of the other body at a function of time will be \({v_o}t\hat i + \left( {{v_0}t\sin \theta - \frac{1}{2}g{t^2}} \right)\hat j\).
Now the distance between the two is simply the euclidean distance given by
\(\sqrt {v_0^2{t^2}{{\cos }^2}\theta + {{\left( {{v_0}t - {v_o}t\sin \theta } \right)}^2}} \)
\( = {v_0}t\sqrt {2\left( {1 - \sin \theta } \right)} \)
\( = 22m\)
Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.
Problem: 1.9
A boat moves relative to water with a velocity which is \(n = 2.0\) times less than the river flow velocity. At what angle to the stream direction must the boat move to minimize drifting?
Solution: 1.9
Here the boats speed is less than that of the river, so there is no way that the boat can move in a direction perpendicular to the flow. Thus there will be a drift. Suppose that the boat travels at an angle \(\theta \) to the perpendicular direction as shown in the figure.
Further suppose that the boat speed is \(v\) and the river speed is \(u\). Then, the boat will travel at a speed \(v\cos \theta \) towards the bank. Thus, if the river is \(x\) units wide it will take \(\frac{x}{{v\cos \theta }}\) time for the boat to reach the other shore. During this entire time however, the boat is drifting with the river with a speed \(\left( {u - v\sin \theta } \right)\). Thus, the drift will be \(\frac{{x\left( {u - v\sin \theta } \right)}}{{v\cos \theta }}\)
The drift will be minimized when,
\(\frac{{d\left( {x\frac{{u - v\sin \theta }}{{v\cos \theta }}} \right)}}{{d\theta }} = 0\)
or, \(\frac{{ - v\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + u\sin \theta }}{{{{\cos }^2}\theta }} = 0\)
or, \(\sin \theta = \frac{v}{u} = \frac{1}{n}\)
That the second derivative is positive at this value indicating that the value obtained in indeed a minima.
Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.
Problem: 1.6
Two boats, \(A\) and \(B\), move away from a buoy anchored at the middle of a river along the mutually perpendicular straight lines. The boat \(A\) along the river, and the boat \(B\) across the river. Having moved off an equal distance from the buoy the boats returned. Find the ration of times of motion of boats \(\frac{{{t_A}}}{{{t_B}}}\) if the velocity of each boat with respect to water is \(\eta = 1.2\) times greater than the stream velocity.
Solution: 1.6
Let us suppose that the boats \(A\) and \(B\) each traveled a distance \(d\) from the buoy before turning back. Let the boats' speed be \(v\) and the river's speed be \(u\).
For \(A\) while going downstream the river aided the boat, thus it moved with a speed \(\left( {v + u} \right)\). On its upstream journey back to the buoy, however the river was opposing the boat and so its speed was \(\left( {v - u} \right)\). The total time taken by \(A\) thus is given by, \({t_A} = \frac{d}{{\left( {v + u} \right)}} + \frac{d}{{\left( {v - u} \right)}}\) ... ... ... ... (i)
Similar to case for \(B\) to travel perpendicular to the buoy, he must move at an angle to the perpendicular to the buoy, he must move at an angle to the perpendicular. His speed perpendicular to the buoy is thus given by, \(v\sqrt {1 - \frac{{{u^2}}}{{{v^2}}}} \).
This will be the same \(B'\) return journey as well. Thus, total time taken by him is given by,
\({t_B} = \frac{{2d}}{{v\sqrt {1 - \frac{{{u^2}}}{{{v^2}}}} }} = \frac{{2d}}{{\sqrt {{v^2} - {u^2}} }}\) ... ... ... ... (ii)
Thus, from (i) and (ii) we get,
\(\frac{{{t_A}}}{{{t_B}}} = \frac{v}{{\sqrt {{v^2} - {u^2}} }} = \frac{\eta }{{\sqrt {{\eta ^2} - 1} }}\)
Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.
Problem: 1.7
Two swimmers leave point \(A\) on one bank of the river to reach point \(B\) lying right across on the other bank. One of them crosses the river along the straight line \(AB\) while the other swims at right angles to the stream and then walks the distance that he has right angles to the stream and then walks the distance that he has been carried away by the stream to get to point \(B\). What was the velocity \(u\) of the walking if both swimmers reached the destination simultaneously? The stream velocity \({v_0} = 2.0km/h\) and the velocity \(v'\) of each swimmer with respect to water equals \(2.5km/h\).
Solution: 1.7
Let us call the swimmer who swam directly from \(A\) to \(B\) as \({S_1}\) and the other swimmer as \({S_2}\). \({S_1}\), in-order to swim directly to \(B\) from \(A\), had to oppose the flow of the river and completely annul it. This means that \({S_1}\) must have had to swim at an angle \(\theta \) from the line \(AB\) such that, the velocity component along the direction of river flow must be exactly equal and opposite to that of the river flow.
\(v'\sin \theta = {v_0}\)
or, \(\sin \theta = \frac{{v'}}{{{v_0}}}\) ... ... ... ... (i)
Thus we have, the component of velocity along the direction \(AB\) of the swimmer \({S_1}\) is given by
\(v'\cos \theta = v'\sqrt {1 - \frac{{v_0^2}}{{v{'^2}}}} \) ... ... ... ... (ii)
If the length \(AB\) is \(d\) then the time taken by swimmer \({S_1}\) to reach \(B\) is given by,
\({t_{{S_1}}} = \frac{d}{{v'\sqrt {1 - \frac{{v_0^2}}{{v{'^2}}}} }}\) ... ... ... ... (iii)
Swimmer \({S_2}\) however, swims perpendicular to the river's flow and hence will be swept off to the right by the river. His speed in the direction perpendicular to the river will be \(v'\) and the river will push him to the right with a speed \({v_0}\). Thus, swimmer \({S_2}\) will take \(\frac{d}{{v'}}\) time to reach the other shore at point \(C\) as seen in the figure. Since, he has been drifting to the right with a speed \(v'\) to the right, \(BC\) will be equal to \(\frac{{d{v_0}}}{{v'}}\). He will thus take \(\frac{{d{v_0}}}{{v'u}}\) time to reach \(B\) from \(C\). This means that the total time swimmer \({S_2}\) takes to reach \(B\) via \(C\) will be equal to,
\({t_{{S_2}}} = \frac{d}{{v'}} + \frac{{d{v_0}}}{{v'u}}\) ... ... ... ... (v)
Since both the swimmers reach at the same time, we have,
\(\frac{d}{{v'\sqrt {1 - \frac{{v_0^2}}{{v{'^2}}}} }} = \frac{d}{{v'}} + \frac{{d{v_0}}}{{v'u}}\)
or, \(u = \frac{{{v_0}}}{{\left[ {\frac{1}{{\sqrt {\left( {1 - \frac{{v_0^2}}{{v{'^2}}}} \right)} }} - 1} \right]}}\)
Here,
\({v_0} = 2km/h\) and \(v' = 2.5km/h\)
Thus, \(u = 3km/h\)
Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.
Problem: 1.6
A ship moves along the equator to the east with velocity \({v_0} = 30km/h\). The southeastern wind blows at an angle \(\phi = 60^\circ \) to the equator with velocity \(v = 15km/h\). Find the wind velocity \(v'\) relative to the ship and the angle \(\phi '\) between the equator and the wind direction in the reference frame fixed to the ship.
Solution: 1.6
If \({v_0}\) is the velocity vector of the ship and \({v_{wind}}\) is the velocity vector of the wind, then the velocity of the wind relative to the ship is simply \(\left( {{v_{wind}} - {v_0}} \right)\).
This is indicated in the above figure. The magnitude \(\left| {{v_{wind}} - {v_0}} \right|\) is then given by:
\( = \sqrt {v_0^2 + v_{wind}^2 + 2{v_0}{v_{wind}}\cos \left( {180^\circ - \phi } \right)} \)
\( = \sqrt {v_0^2 + v_{wind}^2 + 2{v_0}{v_{wind}}\cos \phi } \)
\( = \sqrt {{{15}^2} + {{30}^2} + 2.15.30\cos 60^\circ } \)
\( = \sqrt {225 + 900 + 450} \)
\( = \sqrt {1575} \approx 40km/h\)
The angle of the direction of the wind will be
\(\tan \phi ' = \left( {\frac{{{v_{wind}}\sin \phi }}{{{v_0} + {v_{wind}}\cos \phi }}} \right)\)
\(\tan \phi ' = \left( {\frac{{15\sin 60^\circ }}{{30 + 15\cos 60^\circ }}} \right)\)
\(\phi ' \approx 19^\circ \)
Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.
Problem: 1.5
Two particles, \(1\) and \(2\), moves with constant velocities \({V_1}\) and \({V_2}\). At the initial moment their radius vectors are equal to \({r_1}\) and \({r_2}\). How must these four vectors be interrelated for the particles to collide?
Solution: 1.5
The displacement vector of the first particle as a function of time is \({S_1} = {r_1} + {V_1}t\) and that of the second will be
\({S_2} = {r_2} + {V_2}t\)
Thus,
\({r_1} + {V_1}t = {r_2} + {V_2}t\)
Therefore,
\({r_1} - {r_2} = \left( {{V_2} - {V_1}} \right)t\)
Since \(t\) is only a scalar (1 dimensional) while \(r\) and \(V\) are vector (more than 1 dimensional), for this condition to be true, \(\left( {{r_1} - {r_2}} \right)\) must be aligned in the same direction as \(\left( {{V_2} - {V_1}} \right)\).
Thus,
\(\frac{{\left( {{r_1} - {r_2}} \right)}}{{\left| {{r_1} - {r_2}} \right|}} = \frac{{\left( {{V_2} - {V_1}} \right)}}{{\left| {{V_2} - {V_1}} \right|}}\)
Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.
Problem: 1.4
A point moves rectilinearly in one direction. This figure shows the distance \(S\) traversed by the point as a function of the time \(t\). Using the plot find:
(a) the average velocity of the point during the time of motion.
(b) the maximum velocity.
(c) the time moment \({t_0}\) at which the instantaneous velocity is equal to the mean velocity averaged over the first \({t_0}\) seconds.
Solution: 1.4
(a) The average velocity is distance traveled by time \( = \frac{{200cm}}{{20\sec }} = 10cm/s\)
(b) The maximum velocity is the point where the rate of change of distance that is the slope of the curve is the maximum. In the mid portion of the curve, the point moves \(100cm\) in \(4secs\).
Thus, the slope is \(\frac{{100}}{4}\) = \(25cm/s\). This is the maximum speed achieved.
(c) At any time the instantaneous speed \(\frac{{ds}}{{dt}}\) is the slope of the curve while the average speed \(\frac{s}{t}\) is the tangent of the angle of the line joining the origin to the point.
Let us first consider a time to during the acceleration phase of the point. This is when to is in the interval \(\left( {0,10} \right)\). Throughout this region, the slope of the curve will always be greater than that of the line joining the point to the origin since the curve is convex. This means that throughout the acceleration phase, \(\frac{s}{t}\) will always be less than \(\frac{{ds}}{{dt}}\).
Now lets us consider the region where, the speed is constant. As shown in the picture even in this region the tangent of the line joining the point to the can never catch up the slope of the curve.
However, when we consider a time that is in the decelerating region of the curve, as the instantaneous speed of point decreases, at some time the average speed will catch up with the instantaneous speed. In other words the tangent of angle of line connecting the point to the origin will be same as the slope of the curve. This is shown in the figure beside.
Since exact nature (equation) of the curve of the decelerating part of the curve is not provided (it could be exponentially decaying) it is not possible to mathematically determine the exact value of time at which this happens. However, geometrically one can draw a tangent at every point on the curve and see if it passes through the origin. When this happens it will be the point at which instantaneous speed is same as the average speed.
Here the answer, probably this happens at \(16s\) as suggested answer. Certainly it seems that way but the exact answer cannot be determined mathematically unless the nature of the curve is specified.
Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.
Problem: 1.3
A car starts moving rectilinearly, first with acceleration \(\omega = 5.0m{s^{ - 2}}\) (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate \(\omega \), comes to a stop. The total times of motion equals \(\tau = 25s\). The average velocity during that time is equal to \(\left\langle v \right\rangle = 72km/h\). How long does the car move uniformly?
Solution: 1.3
Mean velocity is total distance by total time. The \(v - t\) graph is shown in the figure above.
Suppose that time of constant acceleration was \(t\). During the initial acceleration phase thus the car will travel \(\frac{1}{2}\omega {t^2}\) and its final speed will be \(\omega t\). This can also be calculated using the \(v - t\) diagram as the area under the line \(AB\) (\(\Delta ABE\)).
The car will take the same time \(t\) to come to complete rest and during this deceleration phase it will travel \(\left( {\omega t} \right)t - \frac{1}{2}\omega {t^2} = \frac{1}{2}\omega {t^2}\).
This can be calculated in the \(v - t\) diagram as the area of the triangle \(\Delta CFD\).
During the uniform motion phase the car travels at a speed \(\omega t\) and travels for the remaining time of \(\left( {\tau - 2t} \right)\).
Thus, during the uniform motion phase it travels a distance of \(\omega t\left( {\tau - 2t} \right)\).
This can be calculated as the area of the rectangle \(EBCF\).
The total distance traveled is thus given by,
\(\frac{1}{2}\omega {t^2} + \omega t\left( {\tau - 2t} \right) + \frac{1}{2}\omega {t^2} = \omega t\tau - \omega {t^2}\)
The average velocity during the entire time is thus given by,
\(\frac{{\omega t\tau - \omega {t^2}}}{\tau } = \left\langle v \right\rangle \)
or, \(\omega {t^2} - \omega t\tau + \left\langle v \right\rangle \tau = 0\)
The above equation is a quadratic equation which has two possible solution,
\(t = \frac{{\omega \tau \pm \sqrt {{\omega ^2}{\tau ^2} - 4\omega \left\langle v \right\rangle \tau } }}{{2\omega }}\)
Clearly we choose the negative sign since \(t\) cannot exceed the total time.
The uniform interval is thus given by,
\(\tau - 2t = \tau \sqrt {1 - \frac{{4\left\langle v \right\rangle }}{{\omega \tau }}} \)
Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.
Problem: 1.2
A point traversed half a distance with a velocity \({V_0}\). The remaining part of the distance was covered with velocity \({V_1}\) for half the time and with velocity \({V_2}\) for the other half of the time. Find the mean velocity of the point average over the whole time of motion.
Solution: 1.2
Mean velocity is total distance by total time. Let the total distance traveled be \(d\). The time taken to travel half the distance \(\left( {\frac{d}{2}} \right)\) at speed \({V_0}\) is \( = \left( {\frac{d}{{2{V_0}}}} \right)\).
Now another \(\left( {\frac{d}{2}} \right)\) distance remains to be traveled.
Now let the time taken to travel this remaining distance be \(t\).
Then, \(\left( {\frac{t}{{2{V_1}}} + \frac{t}{{2{V_2}}}} \right) = \left( {\frac{d}{2}} \right)\)
This means that, \(t = \left( {\frac{d}{{{V_1} + {V_2}}}} \right)\)
The total time traveled thus is, \(\left( {\frac{d}{{2{V_0}}}} \right) + \left( {\frac{d}{{{V_1} + {V_2}}}} \right)\).
Here the total distance is \(d\).
Thus the mean velocity is = \(\left( {\frac{d}{{\frac{d}{{2{V_0}}} + \frac{d}{{\left( {{V_1} + {V_2}} \right)}}}}} \right) = \frac{{2{V_0}\left( {{V_1} + {V_2}} \right)}}{{2{V_0} + {V_1} + {V_2}}}\)
Question: 1.1
A motorboat going downstream overcame a raft at a point \(A\). \(\tau = 60\min \) later it turned back and after some time passed the raft at a distance \(l = 6.0km\) from the point \(A\). Find the flow velocity assuming the duty of the engine to be constant.
Solution:
Since the raft is floating on the river, it moves at the same velocity as the river. Let the velocity of the river be \({V_r}\). Let the velocity of the boat in still water be \(V\). This means that the boat's downstream velocity as seen from the ground will be \(\left( {V + {V_r}} \right)\) and it upstream velocity as seen from the ground will be \(\left( {V - {V_r}} \right)\).
As seen by an observer on the raft, when the motorboat moves upstream, its velocity will be \(\left( {V + {V_r} - {V_r}} \right) = V\) and when its moving downstream, it will be \(\left( { - V + {V_r} - {V_r}} \right) = - V\). Here, positive sign indicates that, the motorboat is moving away from the raft and negative sign indicates that, the motorboat is moving towards the raft.
This figure shows the Time-Displacement diagram of the motorboat and the raft as seen from the point of view of the raft. The slopes of the both upstream and downstream paths of the motorboat relative to the raft are \(V\) and \( - V\). Let \({t_1}\) be the time takes by the raft to meet the boat after it turns around. As seen from the figure, it is obvious that,
\({t_1} = \tau \) ... ... ... (i)
Thus,
The total time between the two meetings of the raft and the boat is \({t_1} + \tau = 2\tau \) ... ... ... (ii)
Now the total distance traveled by the raft is \(l\). This means that,
\(\left( {2\tau } \right) \times {V_r} = l\)
or, \({V_r} = \frac{l}{{2\tau }}\)
or, \({V_r} = \frac{l}{{2\tau }} = \frac{{6.0km}}{{2 \times 60\min }} = \frac{{6.0km}}{{2h}} = 3.0km/h\)
Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.
Physical Fundamentals of Mechanics:
Problem: 1.2
Problem: 1.2
Problem: 1.3
Problem: 1.4
Problem: 1.5
Problem: 1.6
Problem: 1.7
Problem: 1.8
Problem: 1.9
Problem: 1.10
Problem: 1.11
