• Motion in a Plane with Constant Acceleration

• Relative Velocity in Two Dimensions

• Relative Velocity in Two Dimensions

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.

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 ?

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)`

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)|)`

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 ?

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.