Irodov Problem 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 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 towards the bank. Thus, if the river is x units wide it will take time for the boat to reach the other shore. During this entire time however, the boat is drifting with the river with a speed . Thus, the drift will be,
.


This drift will be minimized when,










It is trivial to show that the second derivative is positive at this value indicating that the value obtained is indeed a minima.

Irodov Problem 1.8

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 v+u. On its upstream journey back to the buoy, however, the river was opposing the boat and so its speed was v-u. The total time taken by A thus is given by,





Similar to case of Swimmer 1 in Problem 1.7, for B to travel perpendicular to the buoy, he must move at an angle to the perpendicular. As in Problem 1.7 his speed perpendicular to the buoy is thus given by,. This will be the same B's return journey as well. Thus, the total time taken by him is given by,






Thus, from (1) and (2),

Irodov Problem 1.7

Let us call the swimmer who swam directly from A to B as S1 and the other swimmer as S2.

S1, in-order to swim directly to B from A, had to oppose the flow of the river and completely annul it. This means that S1 must have had to swim at an angle 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.
Thus we have,
Swimmer S1's component along the direction AB is then given by,




If the length if AB is d then the time taken by swimmer S1 to reach B is given by,








Swimmer S2 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 vo. Thus, swimmer S2 will take 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 . He will thus take time to reach B from C. This means that the total time swimmer S2 takes to reach B via C will be equal to,



Since both the swimmers reach at the same time, we have,

Irodov Problem 1.6

If Vo is the velocity vector of the ship and Vwind is the velocity vector of the wind, then the velocity of the wind relative to the ship is simply Vwind - Vo. This is indicated in the figure. The magnitude |Vwind - Vo| is then given by,

 

The angle of the direction of the wind will be,

Irodov Problem 1.5

The displacement vector of the first particle as a function of time is s1 = r1 + t.v1 and that of the second will be s2 = r2+ t.v2. When and if the particles collide, s1 must be equal to s2. Thus,
r1 + t.v1 = r2 + t.v2. Thus, r1-r2 = t.(v2-v1). Since t is only a scalar (1 dimensional) while r and v are vector (more than 1 dimensional), for this condition to be true, r1-r2 must be aligned in the same direction as v2-v1. Thus, (r1-r2)/|r1-r2| = (v2-v1)/|v2-v1|.

Irodov Problem 1.4

a) The average velocity is distance traveled by time = 200cm/20sec = 10cm/sec




b) The maximum velocity is the point where the rate of change of distance i.e. the slope of the curve is the maximum. In the mid portion of the curve, the point moves 100cm in 4 seconds. Thus, the slope is 100/4 = 25cm/sec - this is the maximum speed achieved.

c) At any time the instataneous speed (ds/dt) is the slope of the curve while the average speed (s/t) is the tangent of the angle of the line joining the origin to the point.

Let us first consider a time t0 during the acceleration phase of the point. This is when t0 is in the interval (0,10). 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, s/t will always be less than 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.

Some readers have commented that this probably happens at 16s as suggested in the answer. Certainly it seems that way but the exact answer cannot be determined mathematically unless the nature iof the curve is specified.