Given R=(Asinωt)i+(Acosωt)j
So x=Acosωt => cosωt=xA
y=Asinωt => sinωt=yA
Now we know that
Sin2ωt+cos2ωt=1
So, (xA)2+(yA)2=1
Or, x2+y2=A2
Which is a equation of a circle
Given R=(Asinωt)i+(Acosωt)j
At t=0
R=Ai
At t=2πω
R=Aj
Now average velocity=Displacementt=Aj−Ai2πω=2Aωπ(j−i)
Given R=(Asinωt)i+(Acosωt)j
v=dRdt
=(−Aωsinωt)i+(Aωcosωt)j
a=dvdt
=(−Aω2cosωt)i+(−Aω2sinωt)j
a.v=[(−Aωsinωt)i+(Aωcosωt)j].[(−Aω2cosωt)i+(−Aω2sinωt)j]
=0
We know from above
a=dvdt
=(−Aω2cosωt)i+(−Aω2sinωt)j
=−Aω2R
So it means acceleration directed towards the radial direction
So, tangential acceleration is zero.
Same as above
Radial acceleration= Aω2
Given R=(Asinωt)i+(Acosωt)j
At t=0
R=Ai
At t=πω
R=−Aj
So displacement= −2Ai
Distance =Distance traveled on the half circle = πA
Velocity of A=ui
Velocity of B =−uj
Velocity of B w.r.t to A=Velocity of B-Velocity of A =−u(i+j)
Velocity of A=ui
Velocity of B ==−uj
Velocity of A w.r.t to B=Velocity of A-Velocity of B
=u(i+j)
Position vector of A =uti
Position vector of B=−utj
Relative position vector of particle A wrt to B at time t=Position vector of A - Position vector of B =ut(i+j)
(a) Let us draw the displacement diagram
From the diagram, it is clear for position A
xa=4×(cos45)=2.83 Km
ya=4×(sin45)=2.83 Km
Now taking components of AB vector across horizontal and vertical direction
x=3cos45=2.12 km
y=3sin45=2.12 km
From the figure
xb=xa+x=2.12+2.83=4.95 km
yb=ya−y=2.83−2.12=.71 km
(b) Distance of Point B from the origin is given by
=√x2b+y2b
=5 km
(c) Position vector of point A = xai+yaj=2.83i+2.83j
Position vector of point B =xbi+ybj=4.95i+.71j
(a) and (b)
Horizontal component of velocity
v0x=v0cos40=15.3 m/sec
Vertical component of velocity
v0y=v0sin40=15.3 m/sec
Now let take t be the time in which it strike the wall.
Then
x=v0xt
y=v0yt−gt22
Now x=8 m
So 8=15.3×t
Or t=.52 sec
Now
y=v0yt−gt22
Substituting all the values
Y=5.33 m
(c) Horizontal velocity will remain same during all the motion while vertical velocity will have downward acceleration of g.
So Horizontal component of velocity at the point of strike
vx=15.3 m/sec
Vertical component of velocity at the given point
vy=v0y−gt=12.8−(9.8)(.52)=7.704 m/sec
a. Given
R=(2sin2πt)i+(3cos2πt)j
So co-ordinates of the particles are
x=2sin2πt or sin2πt=x2
y=3cos2πt or cos2πt=y3
Now we know that
(sin2πt)2+(cos2πt)2=1
So
(x2)2+(y3)2=1
Or
x24+y29=1
This is an equation of an ellipse
So the trajectory of the particle is an ellipse
(b) Given
R=(2sin2πt)i+(3cos2πt)j
Velocity vector is given by
v=dRdt
or
v=(4πcos2πt)i+(−6πsin2πt)j
Acceleration vector is given by
a=dvdt
Or
a=(−8π2sin2πt)i+(−12π2cos2πt)j
=−4π2R
Also it indicates that acceleration is towards the center at every point of the motion.
(c)Magnitude of velocity is given by
v=√v2x+v2y
or
v=√16π2cos22πt+36π2sin22πt
=2π√4cos22πt+9sin22πt
=2π√9−5cos22πt
As (sin2πt)2+(cos2πt)2=1
So vmaxwilloccurattimeswhen| cos2 \pi t |=0Or2 \pi t=(2n+1) (\pi /2)wheren=0,1,2.....Ort=1/4,3/4,5/4....Alsovalueofv_{max} =6 \pi Nowv_{min} will occur at times when |cos2πt|=1
Or 2πt=nπ where n=0, 1,2 ......
Or t=0,1/2,1.....
Also value of vmin=4π
(d)
(e)Now when t=0 , cosα=0
or α=90 °
When t=1/4 , cosα=0 or 90°
(a) The situation is depicted the figure below.
Now we know that, the trajectory of the ball is given by
y=(tanθ)x−gx22v20cos2θ
Now here θ=60 so tanθ=√3 and cosθ=12
Or y=(√3)x−2gx2v20 ---(1)
Now the equation of the slope is given by
y=xtanθ
Now here θ=30 so tanθ=1√3
So, y=x√3 ---- (2)
Now the point of strike is the point of intersection of the equation (1) and (2)
Or
x√3=(√3)x−2gx2v20
or
x=v20g√3
Now distance along the slope is given by
x=Scos30
or S=xcos30
=2v203g
(b) Time to reach the point is given by
x=v0xt
=> t=v0cos60x
=> t=2v0g√3
(c) Height above the point O can be found by the equation (2)
y=x√3 ---- (2)
We know that
x=v20g√3
So y=v203g
(d)Horizontal velocity will remain same throughout the motion as there is no acceleration
vx=v0cos60=v0/2
vy=v0y−gt=v0sin60−g×(2v0g√3)=−v02√3
So velocity of strike
v=√v2x+v2y=v0√3
(a) As given in the question velocity vector of particle is
v=ai+bxj
From this we have x and y components of velocity i.e.
and vx=dxdt=a and vy=dydt=bx ------- (1)
From equation 1 we can calculate equation describing motion of particle along x and y axis. Thus integrating for x
∫x0dx=a∫t0dt
or x=at ------ (2)
Now from equation 2 we have
dy=bxdt=batdt
Integrating it we get
∫y0dy=ab∫t0tdt
Or
y=12abt2 --------- (3)
From equations 2 and 3 we get
y=b2ax2 -------- (4)
This is the required equation of particle's trajectory.
(b) Radius of curvature of trajectory y(x) is given as
R=[1+(dydx)2]3/2dydx2 ------ (5)
Differentiating path given in equation 4 for its first and second derivatives we find
and dydx=bax and d2ydx2=ba -------- (6)
Using equation 5 and 6 we find radius of curvature of trajectory as following
R=ab[1+(bax)2]3/2
Let H be the height of the tower
Now equation of trajectory is
y=xtanθ(1−xR)
Now Range is given as a2
when y=H,x=a1
so, H=a1tan45(1−a1a2)=a1(a2−a1)a2
The situation is depicted in fig
v0=33 m/s and θ0=65
The equation of trajectory of the
yb=(tanθ0)x−gx22v20cosθ0
or
yb=2.14x−.025x2
The equation of the incline
yi=(tan28)x=.53x
When the ball strike the incline
yb=yi
or
.53x=2.14x−.025x2
or x=64.4 m
The distance along the incline obeys
x=Scos28 or S=72.9 m
The time to reach that point is given by
x=vcos65t
or
t=4.63 sec
The time taken to reach the pole is given
R=v0cos?t
or t=Rv0cosθ
Now equation of motion of vertical motion
h=v0sinθt−12gt2
at t=Rv0cosθ, h=H
so
H=(v0sinθ)×(Rv0cosθ)−12g(Rv0cosθ2
or H=Rtan?−gR22v20cos2θ
Let us take point A from which the boat departs as the origin of the coordinates system. The direction of the axis are shown below in Figure
The boat moves in a direction perpendicular to the current at constant velocity u.
Let us take a point B on the boat path which is at a distance y from bank. y< b/2
The current velocity will be
v=2v0yb ---(1)
Now time taken to reach the boat at this point B
t=yu
or y=ut
Substituting this value in equation (1)
v=2v0utb
The velocity of the boat with respect to ground
=Velocity of the boat with respect to river + velocity of river with respect to ground
v=2v0utbi+uj
Differentiating it
a=2v0ubi
This will be the acceleration
Now the boat will reach middle point in time
T=b2u
The horizontal distance travelled during this time
s=aT22
Or,
s=v0b4u
Now let take the velocity for y > b/2
The current velocity will be
v=2v0(b−y)b --- (1)
Then
v=2v0(b−y)bi+uj
Differentiating it
a=−2v0ubi
Now the boat will reach bank
T=b2u
The horizontal distance travelled during this time
s=ut+aT22
s=ubu−2v0u2b(bu)2
s=v0b4u
So total distance
s=v0b2u
Now we know from above calculation for y< b/2
y=ut
x=at22
Now a=2v0ub
Substituting this in above
x=v0u0bt2
Now t=yu
So
y2=buv0x
This is a parabola