Physics MOTION IN A PLANE WITH CONSTANT ACCELERATION, RELATIVE VELOCITY
Click for Only Video

Topics Covered

• Motion in a Plane with Constant Acceleration
• Relative Velocity in Two Dimensions

Motion in a Plane with Constant Acceleration

Suppose that an object is moving in x-y plane and its acceleration veca is constant.

Now, let vecv_0= velocity of the object at time t=0

vecv= velocity of the object at time t

Then, by definition

color(green)(veca=(vecv-vecv_0)/(t-0)=(vecv-vecv_0)/t)

Or, color(blue)(vecv=vecv_0 +vecat)

color(red)(ul"In terms of components :")
color(green)(v_x=v_(o x) + a_xt)

color(green)(v_y=v_(oy) + a_yt)

Let us now find how the position vecr changes with time.

Let, color(red)(vecr_0=) position vectors of the particle at time t=0

color(red)(vecr=) position vectors of the particle at time t=t

color(red)(vecv_0=) velocity at time t=0

color(red)(vecv=) velocity at time t

Then, color(red)("Average Velocity" = (vecv+vecv_0)/2)

color(red)("Displacement" =vecr-vecr_0=((vecv+vecv_0)/2) t)=[{(vecv+vecat)+vecv_0}/2]t

vecr-vecr_0=vecv_0t+1/2 vecat^2

color(blue)(vecr=vecr_0+vecv_0t+1/2vecat^2)

color(red)(ul"In component form :")
color(green)(x=x_0+v_(0x)t+1/2a_xt^2)

color(green)(y=y_0+v_(0y)t+1/2a_yt^2)

Now we can conclude that, motion in a plane (two-dimensions) can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions.
A similar result holds for three dimensions.
Q 3065580465

A particle starts from origin at t = 0 with a velocity 5.0 î m//s and moves in x-y plane under action of a force which produces a constant acceleration of (3.0hati + 2.0hatj) m//s^2. (a) What is the y-coordinate of the particle at the instant its x-coordinate is 84 m ? (b) What is the speed of the particle at this time ?

Solution:

The position of the particle is given by

r(t) = V_0 t +1/2 a t^2

= 5.0 hati t +(1/2)(3.0 hati+2.0 hatj) t^2

 = (5.0 t +1.5 t^2) hati +1.0 t^2 hatj

Therefore, x (t ) = 5.0t +1.5t^2
y (t ) = +1.0t^2

Given x (t) = 84 m, t = ?
5.0 t + 1.5 t^2 = 84 ⇒ t = 6 s
At t = 6 s, y = 1.0 (6)^2 = 36.0 m

Now, the velocity V = (dr)/(dt) = (5.0+3.0t) hati+2.0 t hatj

At t = 6 s, v = 23.0hati+12.0hatj

speed | v| = sqrt(23^2+12^2) approx 26 m s^(-1)

Relative Velocity in Two Dimensions

Suppose that two objects A and B are moving with velocities vecv_A and vecv_B (each with respect to some common frame of reference, say ground.).

Then, velocity of object A relative to that of B is
vecv_(AB)=vecv_A-vecv_B

and similarly, the velocity of object B relative to that of A is

vecv_(BA)=vecv_B-vecv_A

Therefore, color(blue)(vecv_(AB) = – vecv_(BA))
and color(blue)(|vecv_(AB)|=|vecv_(BA)|)
Q 3015491360

Rain is falling vertically with a speed of 35 m s^(–1). A woman rides a bicycle with a speed of 12 m s^(–1) in east to west direction. What is the direction in which she should hold her umbrella ?

Solution:

In Fig. v_r represents the velocity of rain and v_b , the velocity of the bicycle, the woman is riding. Both these velocities are with respect to the ground. Since the woman is riding a bicycle, the velocity of rain as experienced by

her is the velocity of rain relative to the velocity of the bicycle she is riding. That is v_(rb) = v_r – v_b

This relative velocity vector as shown in Fig. makes an angle θ with the vertical. It is given by

tantheta = v_b/v_r = 12/35 = 0.343

or theta = 19^0

Therefore, the woman should hold her umbrella at an angle of about 19° with the vertical towards the west.