Acceleration Vector

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

$\mathbf{\bar{a}}=\frac{\Delta \mathbf{v}}{\Delta t}=\mathbf{i} \frac{\Delta v_x}{\Delta t}+\mathbf{j} \frac{\Delta v_y}{\Delta t}$

Or, $\mathbf{\bar{a}}=\bar{a_x}\mathbf{i} +\bar{a_y} \mathbf{j}$

Fig.1: The components vx and vy of velocity v and the angle θ it makes with x-axis. Note that vx = v cos θ, vy = v sin θ.

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

$\boldsymbol{a}=\lim_{\Delta t\rightarrow 0}\frac{\Delta \mathbf{v}}{\Delta t}=\frac{d\mathbf{r}}{dt}$

Since $\Delta \mathbf{v}=\Delta v_x \mathbf{i} +\Delta v_y \mathbf{j}$, we have

$\mathbf{a}= \mathbf{i}\lim_{\Delta t\rightarrow 0} \frac{\Delta v_x}{\Delta t}+\mathbf{j}\lim_{\Delta t\rightarrow 0} \frac{\Delta v_y}{\Delta t}$

      $= \frac{dv_x}{dt}\mathbf{i}+\frac{dv_y}{dt}\mathbf{j}$

$\mathbf{a}=a_x \mathbf{i}+a_y \mathbf{j}$

where $a_x=\frac{dv_x}{dt}$, $a_y=\frac{dv_y}{dt}$

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 Figs. 4.15(a) to (d). P represents the position of the object at time t and P$_1$ , P$_2$ , P$_3$ positions after time ∆t$_1$ , ∆t$_2$ , ∆t$_3$ , respectively (∆t$_1$ > ∆t$_2$ >∆t$_3$). The velocity vectors at points P, P$_1$ , P$_2$ , P$_3$ are also shown in Figs. 4.15 (a), (b) and (c). In each case of ∆t, ∆v is obtained using the triangle law of vector addition. By definition, the direction of average acceleration is the same as that of ∆v. We see that as ∆t decreases, the direction of ∆v changes and consequently, the direction of the acceleration changes. Finally, in the limit ∆t → 0 [Fig. 4.15(d)], the average acceleration becomes the instantaneous acceleration and has the direction as shown.

Fig.2: The average acceleration for three time intervals (a) ∆t$_1$, (b) ∆t$_2$, and (c) ∆t$_3$, (∆t$_1$ > ∆t$_2$ > ∆t$_3$). (d) In the limit ∆t → 0, the average acceleration becomes the acceleration.

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.

Example 1

The position of a particle is given by 

 $\mathbf{r}=3.0t \mathbf{i} +2.0t^2 \mathbf{j}+5.0 \mathbf{k}$

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.

Answer

$\mathbf{v}(t)=\frac{d \mathbf{r}}{dt}=\frac{d}{dt} \left(3.0t \mathbf{i} +2.0t^2 \mathbf{j}+5.0 \mathbf{k} \right)$

       $=3.0 \mathbf{i} +4.0t \mathbf{j}$

$\mathbf{a}(t)=\frac{d \mathbf{v}}{dt}=4.0 \mathbf{j}$

a = 4.0 m.s$^{–2}$ along y- direction 

At t = 1.0 s, $\mathbf{v}=3.0 \mathbf{i} +4.0 \mathbf{j}$

It’s magnitude is

$v=\sqrt{3.0^2+4.0^2}$ = 5.0 m.s$^{–1}$

and direction is

tan θ = $\frac{v_y}{v_x}$, 

θ = tan $^{-1}\left(\frac{4}{3}\right) \cong =53^0$ with x-axis.

Share :