[help] mechanics - newton's law of motion

A particle starts from rest at the point A and travels in a straight line until it reaches the point B.The velocity of the particle t seconds after leaving A is v ms^-1,where v=0.009t^2 -0.0001t^3.Given that the velocity of the particle when it reaches B is zero,find
(¡) the time taken for the particke to travel from A to B
(ii) the distance AB
(iii) the maximum velocity of the particle

Reply 1

Christophicus

aminninja

(i) v = 0,
0.009t^2-0.0001t^3 = 0
t^2(0.009-0.0001t) = 0
t = 0 (at A) or 0.009-0.0001t = 0, t = 0.009/0.0001 = 90 seconds
so time from A to B = 90-0 = 90 seconds

(ii) x =

\int{v} dt

\int0.009t^2 -0.0001t^3dt

Therefore,
AB =

\int_0^{90} 0.009t^2 -0.0001t^3dt

AB = (0.009(90)³/3 - 0.0001(90)⁴/4) - 0
AB = 2187 - 1640.25
AB = 546.75m

(iii) V_max occurs when dv/dt = 0
dv/dt = 2(0.009)t - 3(0.0001)t² = 0
0.018t - 0.0003t² = 0
t(0.018-0.0003t) = 0
t = 0 or t = 0.018/0.0003 = 60 seconds

V_max occurs when t = 60 seconds
V_max = 0.009(60)² - 0.0001(60)³ = 10.8m/s

Reply 2

aminninja

Hey mate am i wrong or it was your mistake.... Arent 0.009/0.0001 =90 ?
My 1st part is t=90 then my 2nd part is 546.75m < i wonder my 1st part is corect or not...

Reply 3

Christophicus

aminninja

Hey mate am i wrong or it was your mistake.... Arent 0.009/0.0001 =90 ?
My 1st part is t=90 then my 2nd part is 546.75m < i wonder my 1st part is corect or not...

Hi. Yeah you're right. I've edited my original post. Can't believe how many mistakes I'm making. It's my own fault for doing little maths/physics over the summer. :frown: