2.7 Graphs of motion with constant acceleration  (Page 5/6)

The position of the particle with respect to origin of reference is given by :

$$x=ut+\frac{1}{2}a{t}^2$$

In order to determine time instants when the particle is at origin of reference, we put x=0,

$$0=15t+\frac{1}{2}X10t^2$$ $$\Rightarrow t^2+3t=0$$ $$\Rightarrow t=0\text{and}-3s$$

Neglecting negative value of time,

$$\Rightarrow t=0$$

The particle is at the origin of reference only at the start of motion. Now, at the point of reversal of direction, speed of the particle is zero. Putting v = 0 and using v=u+at,

$$\Rightarrow 0=15+10t$$ $$\Rightarrow t=-1.5s$$

We neglect negative value and deduce that particle never ceases to move and as such there is no reversal of motion.

The position of the particle with respect to origin of reference is given by :

$$x=x_0+ut+\frac{1}{2}a{t}^2$$

In order to determine time instants when the particle is at origin of reference, we put x=0,

$$\Rightarrow 0=10+15t+\frac{1}{2}X-10X{t}^2$$ $$\Rightarrow -t^2+3t+2=0$$

$$\Rightarrow t=\frac{-3\pm \sqrt{\{9-\left(4X-1X2\right)\}}}{2X-1}$$

$$\Rightarrow t=\frac{-3\pm 4.12}{-2}=-0.56s\text{or}3.56s$$

We neglect negative value of time. The particle reaches origin of reference only once at t = 2 s. In order to determine time instant when particle is at initial position, we put x=10,

$$\Rightarrow 10=10+15t+\frac{1}{2}X-10t^2$$ $$\Rightarrow -t^2+3t=0$$ $$\Rightarrow t=\mathrm{0,}3s$$

The particle is at the initial position twice, including start point. Now, at the point of reversal of direction, speed of the particle is zero. Putting v = 0 and using v=u+at,

$$\Rightarrow 0=15-10t$$ $$\Rightarrow t=1.5s$$

We neglect negative value and deduce that particle never ceases to move and as such there is no reversal of motion.