mechanics

(a) I presume the question ask for the plotting of a velocity-time graph.

The curve is a straight line with +ve slope 20 m/s2 from 0 to 20s, then followed by a straight line parallel to the x-axis from 20s-80s, followed by a staright line of -ve slope to meet the x-axis at 120s. Then a straight line along thex-axis from 120s to 150s. When the train reverses its direction, a straight line with -ve slope of 10 m/s2 below the x-axis from 150s to 160s, followed by a striaght line parallel to the x-axis from 160s to 190s, then followed by a striaght line of +ve slope to meet the x-axis at 200s.

(b)(1) velocity of the train at 20s = 20 x 20 m/s = 400 m/s

Displacement from A to C = area of trapezium unde the graph
= (1/2) x [(60 + 120) x (400)] m = 36 000 m

(2) velocity of the train 10s after starting from C = 10 x 10 m/s = 100 m/s
Displacement from C to B = area of trapezium under the graph
= (1/2) x [(30 + 50) x 100] m = 4000 m

Hence, displacement from A to B = (36000 - 4000) m = 32 000 m

(3) 1st acceleration = 20 m/s2 (given)
2nd acceleration = -400/40 m/s2 = -10 m/s2
3rd acceleration = -10 m/s2 (given)
4th accelration = 100/10 m/s2 = 10 m/s2