Irodov Problem 1.3

Mean velocity is total distance by total time. The v-t graph is shown in the figure beside.

Suppose that time of constant acceleration was t. During the initial acceleration phase thus the car will travel and its final speed will be . This can also be calculated using the v-t diagram as the area under the line AB (triangle ABE).

The car will take the same time t to come to complete rest and during this deceleration phase it will travel . This can be calculated in the v-t diagram as the area of the triangle CFD.

During the uniform motion phase the car travels at a speed and travels for the remaining time of . Thus , during the uniform motion phase it travels a distance of . This can be calculated as the area of the rectangle EBCF.

The total distance traveled is thus given by,



The average velocity during the entire time is thus given by,



The above equation is a quadratic equation which has two possible solutions,


Clearly we choose the -ive sign since t cannot exceed the total time. The uniform interval is thus given by,

Irodov Problem 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 (d/2) at speed vo is d/(2vo). Now another d/2 distance remains to be traveled. Now let the time taken to travel this remaining distance be t. Then (t/2)v1 + (t/2)v2 = d/2. This means that t = d/(v1 + v2).

The total time traveled thus is, d/(2vo) + d/(v1 + v2). The total distance is d. Thus, the mean velocity is,

Irodov Problem 1.1

There are two ways of solving this problem, i) from the point of view of an observer on the raft ii) from the point of view of a person on the ground. I will provide both these methods since, I have got a lot of questions on this problem from the readers.

Since the raft is floating on the river, it moves at the same velocity as the river. Let the velocity of the river be s. 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 v+s and it upstream velocity as seen from the ground will be v-s.

i) Solution From Reference Frame on the Raft

As seen by an observer on the raft, when the boat moves upstream its velocity will be v (v+s-s)  and it will be -v (-v+s-s) when its moving downstream. Here, positive sign means that the boat is moving away from the raft and negative sign means that the boat is moving towards the raft. 

Figure 1 shows the Time-Displacement diagram of the boat and the raft as seen from the point of view of the raft. The slopes of the both upstream and downstream paths of the boat relative to the raft are v and -v.  Let t1 be the time takes by the raft to meet the boat after it turns around. As seen from the figure, it is obvious that



Thus, thus the total time between the two meetings of the raft and the boat is

Now the total distance traveled by the raft is l. This means that,

ii) Solution From Reference Frame on the Ground
The Time-Displacement diagram as seen from the ground of the raft and the boat are shown in Figure 2.

The raft moves at a constant velocity of s downstream (the slope of the line is s) and hence its path is represented by a straight blue line in Figure 2. The boat moves at a speed v+s when  going upstream, represented by a line of slope v+s in Figure 2. The boat turns around and moves at a velocity -v+s when going downstream, represented by a line of negative slope of v-s.

From Figure 2 we can see that,

Now the total distance traveled by the raft is l. This means that,