2.7 Graphs of motion with constant acceleration  (Page 4/6)

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,

$$0=10-15t+\frac{1}{2}X10t^2$$ $$\Rightarrow t^2-3t+2=0$$ $$\Rightarrow t^2-t-2t+2=0$$ $$\Rightarrow t\left(t-1\right)-2\left(t-1\right)=0$$ $$\Rightarrow t=1s\text{and}2s$$

In order to determine time instant when particle is at initial position, we put x=10,

$$\Rightarrow 10=10-15t+\frac{1}{2}X10t^2$$ $$\Rightarrow 5t_2-15t=0$$ $$\Rightarrow t=0\text{and}3s$$

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

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

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,

$$\Rightarrow 0=10+15t+\frac{1}{2}X10t^2$$ $$\Rightarrow t^2+3t+2=0$$ $$\Rightarrow t^2+t+2t+2=0$$ $$\Rightarrow t=-1s\text{and}-2s$$

We neglect both these negative values and deduce that the particle never reaches point of origin. In order to determine time instant when particle is at initial position, we put x=10,

$$\Rightarrow 10=10+15t+\frac{1}{2}X10t^2$$ $$\Rightarrow 5t^2+15t=0$$ $$\Rightarrow t=0\text{and}-3s$$

We neglect negative value. The particle is at the initial position only at the start of motion. Now, at the point of reveral of direction, speed of the particle is zero. Putting v = 0 and using v=u+at,

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

The particle never changes its direction.

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,

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

The particle is at the origin of reference twice, including 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,

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