Home/Archive/Archive - 2015-16/Std-11, Physics, Kinematics, Ch-04 More Problems on 1 D Motion

Std-11, Physics, Kinematics, Ch-04 More Problems on 1 D Motion

So we were dealing with this particular question that the ball is projected vertically upward with some velocity and we are looking forward to find or to calculate distance or displacement covered in last t seconds. What is the meaning of last t seconds? Last t seconds of upward journey, this is what we have written, right. We may have projected this ball at any velocity and there will be a final moment when the velocity will be zero, so final velocity zero. So total displacement during the motion. First let us understand the meaning of last t seconds. What I mean to say is that particle is in upward motion for 10 seconds, that means to reach from here to there it takes 10 seconds. So I am asking you motion of last 3 seconds. I want to know the motion of the last 3 seconds, so last t seconds, some t seconds. Let’s say to reach from here to here it takes capital T time. So from that capital T, how much displacement has it covered in the last t seconds. Done?

Suppose this is the x displacement in the last t seconds, okay. Now how to calculate that?

[Students answering]

Total time, one student is saying that there will be total time capital T and from here to here it will be small t. So what is the time of this journey? Capital T minus small t. If we minus this journey time or displacement from total time so we will get the displacement of last t seconds.

Let’s say, now we don’t know the initial speed. Let’s say initial speed is u, but we don’t know the initial speed. And total time is capital T, if total time is capital T that means after capital T time its velocity will become zero. Yes or on? So will capital T carry this relation or no? There will be this relation between capital T and initial speed. And here we can we can write u as g capital T. Class, is that clear? During the upward journey capital T. Now see we need total, and total we know that if initial speed is u what is the total displacement. Total displacement is equal to maximum height so u square by 2g. And this u square by 2g we can write it with the help of this formula as g square T square divide by 2g, and when we solve that what will we get, gT square by 2. This is total displacement. From total displacement capital T minus small t is the displacement of time. Should we minus that? So this is displacement of Capital T minus small t time. How much is that? Second equation ut plus alpha t square. So it is u, what is the time, capital T minus small t minus g by 2 capital T minus small t square. And again u can be replaced again. It will be gT. Agreed? We can simplify this. This will be gT square minus g capital T small t. We can open this, T square plus small t square minus 2 capital T small t, right. We can simplify it some more. gT square minus g capital T small t s gT square by 2 minus gtsquare by 2 small t. Minus minus becomes plus 2 will be cancelled with 2  g capital T small t.

Here see very carefully, this gets cancelled with one another and this and this can be minused. What is the final value.

2016-03-08T14:20:33+00:00 Categories: Archive - 2015-16|Tags: , , , |0 Comments
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