Question

A particle moves along the x-axis with acceleration a = 6(t- 1), where t is in seconds. If the particle is initially at the origin and moves along the positive x-axis with $v_0$ = 2 m/s, analyze the motion of the particle.

Answer 1

Here we are given a = 6(t- 1) Here, s = +x, v = +v. To find velocity-time relation using a = dv/dt, we have $\frac{dv}{dt}$ = 6 ( t - 1 )...(i)When velocity changes from $v_0$ to v during time t, integrating both sides with respect to time, we have${\int }_{v_0}^{v}dv={\int }_0^{t}6(t-1)dt\Rightarrow v-v_0=3t^2-6t$ Substituting $v_0$ = 2 m/s, we have v = $\frac{dx}{dt}=3t^2-6t+2$ (ii)dx = $(3t^2-6t+2)dt$.(ii) find position-time relation, again integrating (ii) both sides, we have ${\int }_0^{x}dx={\int }_0^t(3t^2-6t+2)dt$ This gives $x=t^3-3t^2$ + 2t = t (t - 1)(t - 2) iii)The particle passes through the origin where x = 0 from (iii), we have t = 0, 1s and 2s. That means the particle crosses the origin twice at t = 1 s and t = 2 s. After t = 2 s, x is positive. Hence, its displacement points in positive x-direction and goes on increasing its magnitude.The particle will come at rest when v = 0. From (ii), we have t = $(1-\frac{1}{\sqrt{3}})$s (=$t_1$, say and t = $(1+\frac{1}{\sqrt{3}})$s (=$t_2$, say)Hence, the particle at t =$t_1$ and $t_2$. That signifies to and fro motion of the particle during first two seconds as it starts retracing its path at t = $t_1$ and t = $t_2$ The displacement (x), velocity (v), and acceleration (a) of the particle in different time intervals are given in the following table and shown in the following graphs.Time interval $(1-\frac{1}{\sqrt{3}})$$(1-\frac{1}{\sqrt{3}})$ $(1+\frac{1}{\sqrt{3}})$