I have an urgent physics question!!!!vectors and motions in a plane?

Question

# A particle starts from the origin at time t = 0 with a velocity of (-6 i hat - 3 j hat)m/s. The particle moves in the x-y plane with a constant acceleration, (2 i hat + 3 j hat) m/s2. At the instant the particle reaches its turning point along the y-axis, what is the magnitude of its displacement from the origin?

Answer 1

Let _ (underscore_ represent vector qualities. also, i and j be unit vectors along +ve x and y axes (x-y plane).

at t=0, Initial velocity U_ = (-6 i - 3 j)
Moves with constant acceleration, f_ = (2 i + 3 j )
At t=t instant, it’s velocity (V_) is given by
f_ = (V_(t) - U_) /t use vectors of U_ & f_
V_(t) = U_ + f_ * t = (-6 i - 3 j) + (2 i + 3 j )*t
V_(t) = i (-6 +2t) + j (- 3 + 3t) …………….(1)

Displacement S_ (t) = U_*t +(1/2)* f_ *t^2 use vectors of U_ & f_
S_ (t) = (-6 i - 3 j)*t +(1/2)* (2 i + 3 j ) *t^2
S_ (t) = i [-6t + t^2] + j [-3t + (3/2)*t^2] ---------(2)

At an instant t=t, particle will reach its turning point along y-axis, when its Vy component will be zero or in V- when coefficient of j unit vector is ZERO

From (1) >> -3 + 3t =0 or t=1 is that instant

V_(t=1) = -4i (moving along x-axis) (turning point along y-axis)
From (2) >> S_ (t=1) = Required = i(-6+1) + j (-3+ (3/2))

S_ (t=1) = i(-5) + j (-1.5)
Its magnitude = |S_| = sqrt (25+2.25)= 5.22 meters

Since the particle has zero displacement at origin, this the magnitude of displacement. (Answer)
------
Analyze: Particle is moving clockwise under f_
Now let us see V_ (1) & S_ (2) at different instants
t=0, V_ = - (6 i + 3 j) , (-x, -y quadrant) S_ = 0
t= 0.5, V_ = - (5 i + 1.5 j) , (-x, -y quadrant) S_ = - (2.75i +1.125j)
t= 0.9, V_ = - (4.2 i + 0.3 j) , (-x, -y quadrant) S_ = - (4.2i +1.125j)
t= 1, V_ = - (4 i ) + 0 ( j ) , (-x, +Y quadrant) S_ = - (5i +1.5j)
Here Vy-goes to different quadrant t=1
-------------------
t= 2, V_ = - (2 i ) + 3 ( j ) , (-x, +Y quadrant)
S_ = - (8i) +0 (j) (changed – Sy displacement turned)
---------------------------------
t= 3, V_ = (0 i ) + 6 ( j ) , (turning point Vx along x, +Y quadrant)
S_ = - (9i) +(4.5j)
So on

Take care: if Vy=0 is taken for y-turning then t=1
If Sy = then t=2

Answer 2

Ok, for me the idea is this :

You know that dv / dt = a

adt = dv

(2hat i + 3hat j)dt = dv >>> now integrate :

2ha*t^2/2 i + 3ha*t^2/2 j = Vf - Vo >> final speed - initial speed

initial speed = -6hat i - 3hat j

ha*t^2 i + 3ha*t^2/2 j - 6ha*t i - 3ha*t j = Vf

Vf = (ha*t^2 - 6ha*t)i + (3ha*t^2 - 3ha*t)j

and : dx / dt = v

vdt = dx

[(ha*t^2 - 6ha*t)i + (3ha*t^2 - 3ha*t)j]dt = dx >> integrate this

(hat^3/3 - 3ha*t^2)i + (ha*t^3 - 3ha*t^2/2)j = Xf - Xo

Xo = initial position = 0,0

then the final position will be :

(hat^3/3 - 3ha*t^2)i + (ha*t^3 - 3ha*t^2/2)j = Xf

this position is like : ai + bj, BUT, THE PROBLEM SAYS THAT THE FINAL POSITION IS ALONG Y AXIS.

Then : ha*t^3/3 - 3ha*t^2 = 0

then t = 9 s

now replace t = 3 at the part y of the position :

Xf = (ha*t^3 - 3ha*t^2/2)j

for t = 9

Xf = 729ha - 729ha/2 = 729ha/2

the magnitude : M = 729ha/2

Hope that might help you