(i) As we observe the graph, we find that the nature of the motion of particles is that, the particles are uniformly accelerated from 0 to 4s and then uniformly retarded from 4s to 6s.
(ii) As we know,
displacement of particles can be obtained by finding the area enclosed by the graph in that part with the time axis up to that instance.
At t = 6 s,
Displacement = area of a triangle
= 1/2 × base × height
= 1/2 × 6 × 2 = 6 m
Hence, Displacement of a particle at t = 6 s is 6 m
(iii) No, the particle does not change its direction of motion.
(iv) At t = 0 to 4 s,
Distance covered = area of the triangle
= 1/2 × base × height
= 1/2 × 4 × 2 = 4 m
Hence, the Distance covered between 0 to 4 s = 4 m
At t = 4 s to 6 s,
Distance covered = area of the triangle
= 1/2 × base × height
= 1/2 × 2 × 2 = 2 m
Hence, Distance covered between 4s to 6 s = 2 m
∴ Distance covered between 0 to 4 s : Distance covered between 4 s to 6 s
= 4 : 2 = 2 : 1
(v) Acceleration in part 0 s to 4 s = slope of the graph
slope = 2-0\over 4-0
slope = 2\over 4 = 0.5 m s-2
Hence, Acceleration in part 0 s to 4 s = 0.5 m s-2
As we know,
Retardation in part 4 s to 6 s = slope of the graph
slope = 0-2\over 6-4
= -2\over 2
= -1 ms-2
Acceleration = -1 ms-2 and as we know retardation is negative acceleration
Hence, Retardation in part 4 s to 6 s = 1 m s-2
