I have been having some troubles with these 2 questions. Not quite sure which way to follow in order to find answer :/?
4. A particle moves on a smooth horizontal plane.
At time t its velocity is given by v = (1 + t2)i + 3tj ms−1.
At t = 4 the position of the particle is r = − i + 4j m.
(a) Find the acceleration at time t = 4.
(b) What was the position at t = 0?
(c) When will it cross the X-axis?
(d) Where will it cross the X-axis?
The answers I found are
a-) dv/dt = 2ti + 3
b-) r= -i + 4j
c-) t=0.70 , 3.79
9. A stone, of mass m, is moving in a straight line along smooth horizontal ground.
At time t, the stone has speed v. As the stone moves, it experiences a total
resistance force of magnitude 2mv3/2. No other horizontal force acts on the
stone. The initial speed of the stone is 9ms−1.
(i) Show that dv/dt = − 2v^(3/2)
(ii) Show that v =9/((3t + 1)^2)
Hint: Put dv/dt=1/(dv/dt) then integrate with respect to v.
(iii) Find the time taken for the speed of the stone to drop to 4ms−1.
question 9-) " As the stone moves, it experiences a total
resistance force of magnitude 2mv^(3/2)
- BrambleLv 76 years agoFavorite Answer
a) a = dv/dt = 2.i+3.j m/s² which is valid for all t
b) s = ∫v.dt = (t+t²).i+c₁.i+(3/2.)t².j+c₂.j
set t=4 : (20+c₁).i+(24+c₂).j = -i+4j giving c₁=-21 and c₂=-20 then:
s = ∫v.dt = (t²+t-21).i+(3t²/2-20).j
If t=0 position s = -21i-20j
c) It crosses the x-axis when the y-coordinate =0 i.e. when 1½.t² =20 or t = √13.33 = 3.65 s
d) Same question as c) so same answer again!
Looking at your question again I think maybe you meant to write t² in the formula for v but you wrote t2 which is the same as 2.t. You should write that t^2. Sorry, you have to take the consequence of your incompetence – that's life! I answer what you ask!
As to your second question ; is it 2m^(3/2) or 2.m³/2? The latter is what you write. Even though it seems dumb to multiply and then divide by 2, I'd have to assume that's what you mean. Clarify it and I'll proceed.