(i) The distance travelled in any part of the graph can be obtained by finding the area enclosed by the graph in that part with the time axis.
(ii) Let E be the point on the time axis (x-axis) at time t and F be the point at time 2t.
When we observe the graph, we find,
Distance travelled in part BC = Area of rectangle EBCF
∴ Distance travelled in part BC = length × breadth
= (2t – t) × (v0 – 0)
= t × v0 —— (1)
Distance travelled in part AB = Area of triangle ABT
∴ Distance travelled in part AB = 1/2 × base × height
= 1/2 × (t – 0) × (v0 – 0)
= 1/2 × t × v0 ——- (2)
Distance travelled in part BC : Distance travelled in part AB
= t × v0 : 1/2 × t × v0
= 1 : 1/2 = 2 : 1
Hence, Distance travelled in part BC : Distance travelled in part AB = 2 : 1
(iii) The different parts of the graph are mentioned below:
(a) Uniform velocity is shown in part BC of the graph, as the velocity is constant with time.
(b) Uniform acceleration is shown in part AB of the graph, as the velocity is increasing with time.
(c) Uniform retardation is shown in the part CD of the graph, as the velocity is decreasing with time.
(iv) (a) The magnitude of the acceleration is lower, as the slope of line AB is less than that of line CD.
(b) Acceleration in part AB = slope of AB
Slope of line AB = v_o-0\over t-0 = v_o\over t
Retardation in part CD = slope of CD
= v_o-0\over 2.5t-2t
= v_o\over 0.5t
Magnitude of acceleration : Magnitude of retardation = Slope of line AB : Slope of line CD
= v_o\over t : v_o\over 0.5t
= 1 : 1\over 0.5
= 1 : 2
Hence, Magnitude of acceleration : Magnitude of retardation = 1 : 2