2.10 Non-uniform acceleration  (Page 5/5)

The particle initially accelerates at a constant rate. It means that its velocity increases with time. The particle, therefore, gains maximum velocity say, $v_{\max }$ , before it begins to decelerate. Let $t_1$ and $t_2$ be the time intervals, for which particle is in acceleration and deceleration respectively. Then,

$$\begin{array}{l}t=t_1+t_2\end{array}$$

From the triangle OAC, the magnitude of acceleration is equal to the slope of straight line OA. Hence,

$$\begin{array}{l}a=\frac{v_{\max }}{t_1}\end{array}$$

Similarly from the triangle CAB, the magnitude of acceleration is equal to the slope of straight line AB. Hence,

$$\begin{array}{l}b=\frac{v_{\max }}{t_2}\end{array}$$

Substituting values of time, we have :

$$\begin{array}{l}t=\frac{v_{\max }}{a}+\frac{v_{\max }}{b}=v_{\max }\left(\frac{a+b}{ab}\right)\end{array}$$

$$\begin{array}{l}{v}_{\max }=\frac{abt}{(a+b)}\end{array}$$

In order to find the displacement, we shall use the fact that area under velocity – time plot is equal to displacement. Hence,

$$\begin{array}{l}x=\frac{1}{2}x{v}_{\max }t\end{array}$$

Putting the value of maximum velocity, we have :

$$\begin{array}{l}\Rightarrow x=\frac{ab{t}^2}{2(a+b)}\end{array}$$

Graphically, slope of the plot yields acceleration and area under the plot yields displacement.

Since the plot involves regular geometric shapes, we shall find areas of these geometric shapes individually and then add them up to get various points for displacement – time plot.

The displacement between times 0 and 1 s is :

$$\begin{array}{l}\Delta x=\frac{1}{2}x1x10=5m\end{array}$$

The velocity between 0 and 1 s is increasing at constant rate. The acceleration is constant and its magnitude is equal to the slope, which is $\raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$1$}\right.=10m/s^2$ . Since acceleration is in the direction of motion, the particle is accelerated. The resulting displacement – time curve is a parabola with an increasing slope.

The displacement between times 1 and 2 s is :

$$\begin{array}{l}\Delta x=1x10=10m\end{array}$$

The velocity between 1 and 2 s is constant and as such acceleration is zero. The resulting displacement – time curve for this time interval is a straight line having slope equal to that of the magnitude of velocity.

The displacement at the end of 2 s is 5 + 10 = 15 m

Now, the displacement between times 2 and 4 s is :

$$\begin{array}{l}\Delta x=\frac{1}{2}x(20+30)x1=25m\end{array}$$

The velocity between 2 and 4 s is increasing at constant rate. The acceleration, therefore, is constant and its magnitude is equal to the slope of the plot, $\raisebox{1ex}{$20$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=10m/s^2$ . Since acceleration is in the direction of motion, the particle is accelerated and the resulting displacement – time curve is a parabola with increasing slope.

The displacement at the end of 4 s is 5 + 10 + 25 = 40 m

Now, the displacement between times 4 and 5 s is :

$$\begin{array}{l}\Delta x=\frac{1}{2}x30x1=15m\end{array}$$

The velocity between 2 and 4 s is decreasing at constant rate, but always remains positive. As such, the acceleration is constant, but negative and its magnitude is equal to the slope, which is $\raisebox{1ex}{$30$}\!\left/ \!\raisebox{-1ex}{$1$}\right.=30m/s^2$ . Since acceleration is in the opposite direction of motion, the particle is decelerated. The resulting displacement – time curve is an inverted parabola with decreasing slope.

The final displacement at the end of 5 s is 5 + 10 + 25 + 15 = 55 m

Characteristics of motion : One dimensional, Unidirectional, variable acceleration in one dimension (magnitude)

In order to answer this question, we need to know the meaning of terminal speed. The terminal speed is the speed when the body begins to move with constant speed. Now, the first time derivate of velocity gives the acceleration of the body :

$$\begin{array}{l}\frac{đv\left(t\right)}{đt}=a=6-3v\left(t\right)\end{array}$$

For terminal motion, a = 0. Hence,

$$\begin{array}{l}a=6-3v\left(t\right)=0\end{array}$$

$$\begin{array}{l}v\left(t\right)=\frac{6}{3}=2m/s\end{array}$$

We have to find the acceleration at t = 0 from the equation given as :

$$\begin{array}{l}\frac{đv\left(t\right)}{đt}=a=6-3v\left(t\right)\end{array}$$

Generally, we would have been required to know the velocity function so that we can evaluate the function for time t = 0. In this instant case, though the function of velocity is not known, but we know the value of velocity at time t = 0. Thus, we can know acceleration at t =0 by just substituting the value for velocity function at that time instant as :

$$\begin{array}{l}a=6-3v\left(t\right)=6-3x0=6m/s^2\end{array}$$

Now when the acceleration is half the initial acceleration,

$$\begin{array}{l}\Rightarrow \mathrm{a’}=\frac{a}{2}=\frac{6}{2}=3m/s^2\end{array}$$

Hence,

$$\begin{array}{l}\Rightarrow 6-3v\left(t\right)=3\\ \Rightarrow v\left(t\right)=\frac{3}{3}=1m/s\end{array}$$