The following question which appeared in AIEEE 2008 question paper is an interesting one. You will find that you can work it out without the knowledge of any standard formula. In fact sheer imagination is sufficient to answer this question and that’s why it is interesting. Here is the question:
A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x-direction with a constant speed. The position of the first body is given by x1(t) after time 't' and that of second body by x2(t) after the same time interval. Which of the following graphs correctly describes (x1 – x2) as a function of time ‘t’?
(x1 – x2) must be negative in the initial stage since the first body starts from rest where as the second body has a constant speed through out its motion. The first body will be behind the second one and the separation between them will increase for some time. As the first body picks up speed, the separation between the bodies decreases and becomes zero at a certain instant. The separation (x1 – x2) will be negative up to this instant. Thereafter the first body overtakes the second one and the separation (x1 – x2) becomes positive and goes on increasing. The situation is indicated correctly by graph (1).
Alternatively, you can find the answer by applying the equation,
s = ut + (1/2)at2 for uniformly accelerated one dimensional motion.
Thus x1 = 0 + (1/2)at2 and x2 = vt where a is the constant acceleration of the first body and v is the constant speed of the second body.
Therefore, (x1 – x2) = (1/2)at2 – vt
(x1 – x2) = 0 when t = 0 and when t = 2v/a.
Further, d(x1 – x2)/dt = at – v so that at t = v/a the quantity (x1 – x2) will be a maximum or minimum. It is in fact a minimum point on the curve obtained by plotting (x1 – x2) against t, as shown by the positive value of the second differential coefficient, d2(x1 – x2)/dt2.
So the correct option is indeed curve (1).
Here is another question:
A particle starts from rest with a constant acceleration and moves along the positive x-direction. If its displacement is x1 in the first two seconds and x2 in the next two seconds, then
(a) x2 = 2x1
(b) x2 = 3x1
(c) x2 = 4x1
(d) x2 = 5x1
(e) x2 = 6x1
We have x1 = (1/2) a×22 = 2a
x2 = (1/2) a×42 – (1/2) a×22 = 6a
Therefore, x2 = 3x1
Now, consider the following question which is quite simple. But be careful while answering.
A train having a constant speed of 54 km/hour takes 10 seconds to move past a lamp post. How much time is required for this train to cross a 300 m long bridge?
(a) 18 s
(b) 20 s
(c) 30 s
(d) 36 s
(e) 40 s
The speed of the train in ms–1 is 54×5/18 = 15.
Since the train takes 10 s to move past the lamp post, the length of the train is 15×10 = 150 m.
The train has to travel a total distance of 450 m to cross the bridge. (This includes the length of the train also).
Therefore, time required = 450/15 = 30 s.
The following question appeared in AIPMT 2008 question paper:
A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms–1 to 20 ms–1 while passing through a distance 135 m in t second. The value of t is
(1) 1.8
(2) 12
(3) 9
(4) 10
We have vt 2 = v0 2 + 2as with usual notations.
Therefore, 202 = 102 + 2a×135 from which a = 30/27
Substituting this value of acceleration a in the equation
vt = v0 + at
we have 20 = 10 + (30/27)t from which t = 9 second
You will find some useful posts on one dimensional motion at physicsplus
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