Please Wait... While Loading Full Video#### CLASS 11 CHAPTER 4 - MOTION IN A PLANE

### MOTION IN A PLANE

• Motion in a Plane

• Position Vector

• Displacement

• Velocity

• Acceleration

• Position Vector

• Displacement

• Velocity

• Acceleration

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Position Vector")} `

The position vector `vecr` of a particle P located in a plane with reference to the origin of an x-y reference frame (Fig. a) is given by

`color(blue)(vecr = x hati+ yhatj)`

where `color(blue)(x)` and `color(blue)(y)` are components of `color(blue)(r)` along x-, and y- axes or simply they are the coordinates of the object.

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Displacement Vector")} `

Suppose a particle moves along the curve shown by the thick line and is at `P` at time `t` and `P′` at time `t′ `(Fig. b). Then, the displacement is :

`color(blue)(Deltavecr=vecr^' -vecr)` and is directed `color(green)("from P to P′")`

We can write in a component form

`Deltavecr=(x^' hati + y^' hatj)-(x hati+yhatj)`

`color(red)(Deltavecr=hatiDeltax+hatjDeltay)`

where `color(red)(Deltax=x^' -x, \ \ \ \ Deltay=y^'-y)`

The position vector `vecr` of a particle P located in a plane with reference to the origin of an x-y reference frame (Fig. a) is given by

`color(blue)(vecr = x hati+ yhatj)`

where `color(blue)(x)` and `color(blue)(y)` are components of `color(blue)(r)` along x-, and y- axes or simply they are the coordinates of the object.

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Displacement Vector")} `

Suppose a particle moves along the curve shown by the thick line and is at `P` at time `t` and `P′` at time `t′ `(Fig. b). Then, the displacement is :

`color(blue)(Deltavecr=vecr^' -vecr)` and is directed `color(green)("from P to P′")`

We can write in a component form

`Deltavecr=(x^' hati + y^' hatj)-(x hati+yhatj)`

`color(red)(Deltavecr=hatiDeltax+hatjDeltay)`

where `color(red)(Deltax=x^' -x, \ \ \ \ Deltay=y^'-y)`

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Average Velocity")} `

The average velocity (`barv`) of an object is the ratio of the displacement and the corresponding time interval :

`color(blue)(barv=(Deltavecr)/(Deltat))=(Deltaxhati+Deltayhatj)/(Deltat)=hati(Deltax)/(Deltat)+hatj(Deltay)/(Deltat)`

`color(red)(barv=barv_xhati+barv_yhatj)`

The direction of the average velocity is the same as that of `Δvecr`.

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Instantaneous Velocity")} `

The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero :

`color(blue)(vecv=lim_(Deltat->0) (Deltavecr)/(Deltat)=(dvecr)/(dt))`

The meaning of the limiting process can be easily understood with the help of `color(red)("Fig (a) to (d).")`

In these figures, the thick line represents the path of an object, which is at P at time t. `P_1`, `P_2` and `P_3` represent the positions of the object after times `Δt_1,Δt_2,` and `Δt_3`. `Δr_1, Δr_2,` and `Δr_3` are the displacements of the object in times `Δt_1, Δt_2`, and `Δt_3`, respectively.

The direction of the average velocity `barv` is shown in figures (a), (b) and (c) for three decreasing values of Δt, i.e. `Δt_1,Δt_2`, and `Δt_3, (Δt_1 > Δt_2 > Δt_3)`. As Δt → 0, Δr → 0 and is along the tangent to the path (Fig. d).

Therefore, the direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion.

We can express `vecv` in a component form

`color(red)(vecv=(dvecr)/(dt))`

`=lim_(Deltat->0) [(Deltax)/(Deltat) hati + (Deltay)/(Deltat)hatj]`

`=hatilim_(Deltat->0)(Deltax)/(Deltat) + hatjlim_(Deltat->0)(Deltay)/(Deltat)`

`color(blue)(vecv=hati(dx)/(dt)+hatj(dy)/(dt))`

where `v_x=(dx)/(dt), v_y=(dy)/(dt)`

The magnitude of `vecv` is then `v=sqrt(v_x^2+v_y^2)`

and the direction of `vecv` is given by the angle `θ :`

`color(red)(tan theta=(v_y)/(v_x), \ \ \ \ theta=tan^(-1)[(v_y)/(v_x)])`

The average velocity (`barv`) of an object is the ratio of the displacement and the corresponding time interval :

`color(blue)(barv=(Deltavecr)/(Deltat))=(Deltaxhati+Deltayhatj)/(Deltat)=hati(Deltax)/(Deltat)+hatj(Deltay)/(Deltat)`

`color(red)(barv=barv_xhati+barv_yhatj)`

The direction of the average velocity is the same as that of `Δvecr`.

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Instantaneous Velocity")} `

The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero :

`color(blue)(vecv=lim_(Deltat->0) (Deltavecr)/(Deltat)=(dvecr)/(dt))`

The meaning of the limiting process can be easily understood with the help of `color(red)("Fig (a) to (d).")`

In these figures, the thick line represents the path of an object, which is at P at time t. `P_1`, `P_2` and `P_3` represent the positions of the object after times `Δt_1,Δt_2,` and `Δt_3`. `Δr_1, Δr_2,` and `Δr_3` are the displacements of the object in times `Δt_1, Δt_2`, and `Δt_3`, respectively.

The direction of the average velocity `barv` is shown in figures (a), (b) and (c) for three decreasing values of Δt, i.e. `Δt_1,Δt_2`, and `Δt_3, (Δt_1 > Δt_2 > Δt_3)`. As Δt → 0, Δr → 0 and is along the tangent to the path (Fig. d).

Therefore, the direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion.

We can express `vecv` in a component form

`color(red)(vecv=(dvecr)/(dt))`

`=lim_(Deltat->0) [(Deltax)/(Deltat) hati + (Deltay)/(Deltat)hatj]`

`=hatilim_(Deltat->0)(Deltax)/(Deltat) + hatjlim_(Deltat->0)(Deltay)/(Deltat)`

`color(blue)(vecv=hati(dx)/(dt)+hatj(dy)/(dt))`

where `v_x=(dx)/(dt), v_y=(dy)/(dt)`

The magnitude of `vecv` is then `v=sqrt(v_x^2+v_y^2)`

and the direction of `vecv` is given by the angle `θ :`

`color(red)(tan theta=(v_y)/(v_x), \ \ \ \ theta=tan^(-1)[(v_y)/(v_x)])`

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Average Acceleration")} `

The average acceleration `bara` of an object for a time interval Δt moving in x-y plane is the change in velocity divided by the time interval :

`color(blue)(bara=(Deltavecv)/(Deltat))=(Deltav_xhati+Deltav_yhatj)/(Deltat)=(Deltav_x)/(Deltat)hati+(Deltav_y)/(Deltat)hatj`

`color(red)(bara=a_xhati+a_yhatj)`

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Instantaneous Acceleration")} `

The acceleration (instantaneous acceleration) is the limiting value of the average acceleration as the time interval approaches zero :

`color(blue)(veca=lim_(Deltat->0) (Deltavecv)/(Deltat))`

Since `Δvecv = Δv_xhati + Δv_yhati`, we have

`veca=hati lim_(Deltat->0)[(Deltav_x)/(Deltat)] + hatj lim_(Deltat->0)[(Deltav_y)/(Deltat)]`

`veca=a_xhati+a_yhatj`

where `a_x=(dv_x)/(dt), a_y=(dv_y)/(dt)`

As in the case of velocity, we can understand graphically the limiting process used in defining acceleration on a graph showing the path of the object’s motion. This is shown in `color(red)("Figs. (a) to (d).")`

`P` represents the position of the object at time t and `P_1, P_2, P_3` positions after time `Δt_1, Δt_2, Δt_3`, respectively (`Δt_1 > Δt_2 > Δt_3`). The velocity vectors at points `P, P_1, P_2, P_3` are also shown in Figs. (a), (b) and (c). In each case of Δt, Δv is obtained using the triangle law of vector addition. By definition, the direction of average acceleration is the same as that of `Δv.` We see that as `Δt` decreases, the direction of `Δv` changes and consequently, the direction of the acceleration changes. Finally, in the limit `Δt->0` Fig. (d), the average acceleration becomes the instantaneous acceleration and has the direction as shown.

`\color{green} ✍️ \color{green} \mathbf(KEY \ CONCEPT)`

Note that in one dimension, the velocity and the acceleration of an object are always along the same straight line (either in the same direction or in the opposite direction). However, for motion in two or three dimensions, velocity and acceleration vectors may have any angle between `0°` and `180°` between them.

The average acceleration `bara` of an object for a time interval Δt moving in x-y plane is the change in velocity divided by the time interval :

`color(blue)(bara=(Deltavecv)/(Deltat))=(Deltav_xhati+Deltav_yhatj)/(Deltat)=(Deltav_x)/(Deltat)hati+(Deltav_y)/(Deltat)hatj`

`color(red)(bara=a_xhati+a_yhatj)`

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Instantaneous Acceleration")} `

The acceleration (instantaneous acceleration) is the limiting value of the average acceleration as the time interval approaches zero :

`color(blue)(veca=lim_(Deltat->0) (Deltavecv)/(Deltat))`

Since `Δvecv = Δv_xhati + Δv_yhati`, we have

`veca=hati lim_(Deltat->0)[(Deltav_x)/(Deltat)] + hatj lim_(Deltat->0)[(Deltav_y)/(Deltat)]`

`veca=a_xhati+a_yhatj`

where `a_x=(dv_x)/(dt), a_y=(dv_y)/(dt)`

As in the case of velocity, we can understand graphically the limiting process used in defining acceleration on a graph showing the path of the object’s motion. This is shown in `color(red)("Figs. (a) to (d).")`

`P` represents the position of the object at time t and `P_1, P_2, P_3` positions after time `Δt_1, Δt_2, Δt_3`, respectively (`Δt_1 > Δt_2 > Δt_3`). The velocity vectors at points `P, P_1, P_2, P_3` are also shown in Figs. (a), (b) and (c). In each case of Δt, Δv is obtained using the triangle law of vector addition. By definition, the direction of average acceleration is the same as that of `Δv.` We see that as `Δt` decreases, the direction of `Δv` changes and consequently, the direction of the acceleration changes. Finally, in the limit `Δt->0` Fig. (d), the average acceleration becomes the instantaneous acceleration and has the direction as shown.

`\color{green} ✍️ \color{green} \mathbf(KEY \ CONCEPT)`

Note that in one dimension, the velocity and the acceleration of an object are always along the same straight line (either in the same direction or in the opposite direction). However, for motion in two or three dimensions, velocity and acceleration vectors may have any angle between `0°` and `180°` between them.

Q 3045380263

The position of a particle is given by `r = 3.0t hati + 2.0t^2hatj + 5.0 hatk` where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find `v(t)` and a(t) of the particle. (b) Find the magnitude and direction of `v(t)` at `t = 1.0 s.`

`v(t) = (dr)/(dt) = d/(dt) (3.0 t hati+2.0 t^2 hatj+5.0 hatk)`

` = 3.0 hati+4.0 t hatj`

`a(t) = (dv)/(dt) = +4.0 hatj`

`a = 4.0 ms^(-2)` along y- direction At `t = 1.0 s, v = 3.0 hati + 4.0hatj`

It’s magnitude is `v = sqrt(3^2+4^2) = 5.0 m s^(-1)` and direction is

`theta = tan^(-1) (v_y/v_x) = tan^(-1) (4/3) approx 53^0` with x-axis.