motion qu (1 Viewer)

[redslert]

a particle starts to move at 0 the origin with an initial velocity of 20cm/s. its acceleration at the end of 't' seconds after commencement of its motion is:

a=18-2t (cm/s^2)

find:
(i) the velocity of the particle at the end of 3sec
(ii) its distance from 0 at the end of 3sec

i got part (i) which is 65cm/s

but i cant work out part (ii)
help?

 

[McLake]

(i)
a = 18-2t
v = 18t - t^2 + C
when t=0, v=20
so v = 18t - t^2 + 20

for v when t=3 sub in 3
v = 65 cm/s


(ii)
we know v = 18t - t^2 + 20
so x = 9t^2 - 1/3t^3 + 20t + C
when t=0, x=0
so x = 9t^2 - 1/3t^3 + 20t

for x when t=3 sub in 3
x = 132 cm

 

[redslert]

ar thankz!

i feel so stupid now
forgot about the 20 when intergrating for S or u wrote (x)