2.7 Graphs of motion with constant acceleration  (Page 6/6)

$$\begin{array}{l}a=\frac{đv}{đt}=-10m/s^2\end{array}$$

Alternatively, we can see that expression of velocity has the form v = u + at. Evidently, the motion has a constant acceleration of -10 $m/s^2$ . Also, note that origin of reference and initial position are same. Applying equation of motion for position as :

$$\begin{array}{l}x=ut+\frac{1}{2}a{t}^2=40t+\frac{1}{2}x-10x{t}^2=40t-5t^2\end{array}$$

According to question, we have to find the time when particle is 60 m away from the initial position. Since it is one dimensional motion, the particle can be 60 m away either in the positive direction of the reference or opposite to it. Considering that it is 60 m away in the positive reference direction, we have :

$$\begin{array}{l}60=40t-5t^2\end{array}$$

Re-arranging,

$$\begin{array}{l}{t}^2-8t+12=0\\ t=2s\mathrm{or}6s\end{array}$$

We observe here that the particle is ultimately moving in the direction opposite to reference direction. As such it will again be 60 away from the initial position in the negative reference direction. For considering that the particle is 60 m away in the negative reference direction,

$$\begin{array}{l}-60=40t-5t^2\end{array}$$

$$\begin{array}{l}\Rightarrow t^2-8t-12=0\\ t=\frac{-(-8)\pm \surd \{{(-8)}^2-4x1x(-12\left)\right\}}{2}=-1.29s,9.29s\end{array}$$

Neglecting negative value of time,

$$\begin{array}{l}\Rightarrow t=9.29s\end{array}$$

Note : We can also solve the quadratic equation for zero displacement to find the time for the particle to return to initial position. This time is found to be 8 seconds. We note here that particle takes 2 seconds to reach the linear distance of 60 m in the positive direction for the first time, whereas it takes only 9.29 – 8 = 1.29 second to reach 60 m from the initial position in the negative direction.

The particle is decelerated while moving in positive direction as velocity is positive, but acceleration is negative. On the other hand, both velocity and acceleration are negative while going away from the initial position in the negative direction and as such particle is accelerated. Therefore, the particle takes lesser time to travel same linear distance in the negative direction than in the positive direction from initial position..

Hence, the particle is 60 m away from the initial position at t = 2 s, 6 s and 9.29 s.

t (s) 0 2 4 6 8 9.29 x (m) 0 60 80 60 0 -60v (m/s) 40 20 0 -20 -40 -52.9