2.10 Non-uniform acceleration  (Page 3/5)

Problem : The motion of a body in one dimension is given by the equation dv(t)/dt = 6.0 -3 v (t), where v(t) is the velocity in m/s and “t” in seconds. If the body was at rest at t = 0, then find the expression of velocity.

Solution : Here, acceleration is given as a function in velocity as independent variable. We are required to find expression of velocity in time. Using integration result obtained earlier for expression of time :

$$\Rightarrow \text{Δ}t=\int \frac{đv}{a\left(v\right)}$$

Substituting expression of acceleration and integrating between t=0 and t=t, we have :

$$\begin{array}{l}\Rightarrow \text{Δ}t=t=\int \frac{đv\left(t\right)}{6-3v\left(t\right)}\end{array}$$

$$\begin{array}{l}\Rightarrow t=\frac{ln\left[\frac{\{6-3v(t)}{6}\right]}{-3}\end{array}$$

$$\begin{array}{l}\Rightarrow 6-3v\left(t\right)=6e^{-3t}\\ \Rightarrow 3v\left(t\right)=6(1-e^{-3t})\\ \Rightarrow v\left(t\right)=2(1-e^{-3t})\end{array}$$

Problem : A particle moving in a straight line is subjected to accelerations as given in the figure below :

If v = 0 and t = 0, then draw velocity – time plot for the same time interval.

Solution : In the time interval between t = 0 and t = 2 s, acceleration is constant and is equal to 2 $m/s^2$ . Hence, applying equation of motion for final velocity, we have :

$$\begin{array}{l}v=u+at\end{array}$$

But, u = 0 and a = 2 $m/s^2$ .

$$\begin{array}{l}\Rightarrow v=2t\end{array}$$

The velocity at the end of 2 seconds is 4 m/s. Since acceleration is constant, the velocity – time plot for the interval is a straight line. On the other hand, the acceleration is zero in the time interval between t = 2 and t = 4 s. Hence, velocity remains constant in this time interval. For time interval t = 4 and 6 s, acceleration is - 4 $m/s^2$ .

From the plot it is clear that the velocity at t = 4 s is equal to the velocity at t = 2 s, which is given by :

$$\begin{array}{l}\Rightarrow u=2x2=4m/s\end{array}$$

Putting this value as initial velocity in the equation of velocity, we have :

$$\begin{array}{l}\Rightarrow v=4-4t\end{array}$$

We should, however, be careful in drawing velocity plot for t = 4 s to t = 6 s. We should relaize that the equation above is valid in the time interval considered. The time t = 4 s from the beginning corresponds to t = 0 s and t = 6 s corresponds to t = 2 s for this equation.

Velocity at t = 5 s is obtained by putting t = 1 s in the equation,

$$\begin{array}{l}\Rightarrow v=4-4x1=0\end{array}$$

Velocity at t = 6 s is obtained by putting t = 2 s in the equation,

$$\begin{array}{l}\Rightarrow v=4-4x2=-4m/s\end{array}$$

Problem : A particle starting from rest and undergoes a rectilinear motion with acceleration “a”. The variation of “a” with time “t” is shown in the figure. Find the maximum velocity attained by the particle during the motion.

Solution : We see here that the particle begins from rest and is continuously accelerated in one direction (acceleration is always positive through out the motion) at a diminishing rate. Note that though acceleration is decreasing, but remains positive for the motion. It means that the particle attains maximum velocity at the end of motion i.e. at t = 12 s.

The area under the plot on acceleration – time graph gives change in velocity. Since the plot start at t = 0 i.e. the beginning of the motion, the areas under the plot gives the velocity at the end of motion,

$$\begin{array}{l}\Delta v=v_2-v_1=v-0=v=\frac{1}{2}x8x12=48m/s\end{array}$$

$$\begin{array}{l}{v}_{\max }=48m/s\end{array}$$