Thursday, February 26, 2009

Q) A particle of mass m moves on the x-axis as follows: it starts from rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times ( 0 < t < 1).

If α denotes the magnitude of instantaneous acceleration of the particle, then which of the following is/are true?
(a) α cannot remain positive throughout
(b) α cannot exceed 2 at any point in its path
(c) α must be equal to 4 at some point or points in its path
(d) α must change sign during the motion, but no other assertion can be made with the information given

(a) is obviously true, since for motion to initiate and end, acceleration and deceleration must be provided.

Now, for determining the least value of maximum acceleration in this motion, the particle can be considered to be moving in SHM (simple harmonic motion) with equilibrium position x = 1/2 and T (time period) = 2 seconds.

acceleration = -w^2 .x

Max acceleration
in such a motion will be = w^2 . Amplitude = (2pi/2)^2 . (1/2) = pi^2 / 2 = 4.934 m/s^2

And max. acceleration would be higher if any other type of motion occurred (instead of SHM). So the least value of max. acceleration is 4.934 m/s^2 which means that 4m/s^2 acceleration HAS to occur at some point in the path of the particle's motion. (which implies that (c) is true too).

Hence, (a) and (c) are true.



meeoww said...

thts a gud work on your side .provideing a helping hand for others . keepup the gud work. by the way . wht course are yu in in usa?

spideyunlimited said...

Undergraduate. Computer Science and Math.