# need a little help... (1 Viewer)

#### RHINO7

##### Member

i) A particle has a constant accelerationof 12ms^-2. If the particle has a velocity of 2ms^-1 and is 3m from then origin after 5s, find its displacement after 10s.

ii) A particle is accelerating according to the equation a=(3t+1)^2 ms^-2. If the particle is initially at rest 2m to the left of the 0, find its displacement after 4s.

#### webby234

##### Member
a = 12
v = 12t + c
at t = 5 v = 2 so c = -58
v = 12t - 58
x = 6t2 - 58t + c
t = 5 x = 3
c = 143
x = 6t2 - 58t + 143
Sub in t = 10

ii) same sort of thing - you have the information that at t = 0 v = 0 x = -2 which you can use to find the constants of integration.

#### Pastafarian

##### New Member
I) a= 12
Integrating v= 12t + c
2=12(5) + c
c=-56

v=12t - 56
integrate again
x= 6t^2 -56t + c
3=6(5)^2 - 56(5) + c
c= 133

x= 6t^2 -56t + 133
t=10 x= 173

edit: go my arithmetic

#### RHINO7

##### Member
sorry i'm having trouble with this particular question---

i) The Velocity of a particle is given by v= 4cos2t ms^-1. If the particle is 3m to the right of the origin after pi s, find the exact

a) displacement after pi/6 s-

b) acceleration after pi/6 s-

#### zingerburger

##### Member
a) Integrate v=4cos2t.
That will be x=2sin2t + k metres.
When x=3 and t=pi:
3=2sin(2pi) + k
Therefore k=3.
When t=pi/6:
x=2sin(2pi/6) + 3
=2sin(pi/3) + 3
=root3 + 3 metres

b) Differentiate v=4cos2t.
That will be a=-8sin2t.
When t=pi/6:
a=-8sin(2pi/6)
=-8sin(pi/3)
=-4root2 ms^-2