Question

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)

Answer 1

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

Answer 2

(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

Answer 3

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.

Answer 4

i made a mistake with displsment from CtoB

Answer 5

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