6.1 Displacement - time graph

1.

Body is stationary

Slope is $0$, $$\frac{{ds}}{{dt}} = 0$$ So, $v=0$

Also, it is a straight line. $$\frac{{{d^2}s}}{{d{t^2}}} = 0$$ So, $$a = 0$$

2.

Uniform motion

As slope is constant. So, $$v = \frac{{ds}}{{dt}}$$$$v = {\text{constant}}$$

Also, it is a straight line. So, $$\frac{{{d^2}s}}{{d{t^2}}} = 0$$$$a = 0$$

3.

Uniformly acceleration motion

As we can see that slope changes at every point in a curve. So, velocity is variable.

$$v = \frac{{ds}}{{dt}} = {\text{variable}}$$

As the curve is concave up. So, the acceleration is positive. $$a = \frac{{{d^2}s}}{{d{t^2}}} > 0$$

If acceleration is positive, velocity will increase.

4.

Uniformly retarded motion

As we can see that slope changes at every point in a curve. So, velocity is variable.

$$v = \frac{{ds}}{{dt}} = {\text{variable}}$$

As the curve is concave down. So, the acceleration is negative. $$a = \frac{{{d^2}s}}{{d{t^2}}} < 0$$

If acceleration is negative, velocity will decrease.

5.

Uniform motion

e$$v = \frac{{ds}}{{dt}} = {\text{constant}}$$$$\tan \theta = - ve$$

So, velocity is in negative direction or we can say it is moving towards the observer.

Acceleration is $0$, as the curve is a straight line. $$\frac{{{d^2}s}}{{d{t^2}}} = 0$$ So, $$a = 0$$