Question

Please help me with this question

Answer 1

Explanation:

a) the object executes 'constant velocity' in part AB of the graph.

b) acceleration in BC=∆v/∆t={(0-20)/(40-30)}m/s2= -10 m/s2

c) Distance=area of trap. AOCB={1/2*(AB+OC)*AO}= (1/2*(20+40)*20)m

= (1/2*60*20)m=600 m