A particle moves in a straight line from A to C, passing through B. It starts at rest at A and accelerates uniformly at 1ms/-2 for 5s before arriving at B. Between B and C it then accelerates uniformly at 2ms/-2 for another 5s. What is the average speed of the particle from its journey from A (Mechanics)

take out the total distance using la of motion. and divide it by total time.

(dis)1=1/2at^2 where a is accl. and t is time. =25/2 (dis)2=2*25/2=25 total distance=25+25/2 divide it by 10

From A to B: u=0, a=1 and t-5. So to find the distance we will use the formula: \[s_{1}=ut+(1/2)at^2\] sub in the values and find s1. s1 is the distance from A to B. From B to C: we first need to find the velocity of the object once reached B. we will use the formula: \[v=u+at\] v is the velocity of the object at B. okay now continue doing B to C. now our initial velocity is v(we found earlier). a=2,t=5 By using the same formula: \[s_2=vt+(1/2)at^2\] Find s2 and the total distance S=s1 + s2. average speed=S/Total time. total time would be 10s.

i made a mistake with displsment from CtoB

human make mistakes, that's part of growing up. XD