KinematicsEasy
Question
A particle of mass m moves on the 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 instantaneous accel-eration of the particle, then
Options
A.α can not remain positive for all t in the interval 0 ≤ t ≤ 1
B.|α| cannot exceed 2 at any point in its path
C.|α| must be ≥ 4 at some point or points in its path
D.α must change sign during the motion, but noo other assertion can be made with the information given
Solution
Since, the body is at rest at x = 0 and x = 1, α cannot be positive for all in the time interval 0 ≤ t ≤ 1.
Therefore, first the particle is accelerated and then retarded.
Now, total time t = 1 s (given)
total displacement s = 1 m (given)
s = Area under v-t graph
∴ Height or vmax =
= 2m/s is also fixed.[Area or s =
× t × vmax]If the height and base are fixed, area is also fixed
In case 2 : Acceleration = Retardation = 4 m/s2
In case : 1 Acceleration > 4 m/s2 while
Retardation < 4 m/s2
While in case : 3 Acceleration < 4 m/s2 and
Retardation > 4 m/s2
Hence, |α| ≥ 4 at some point or points in its path.
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