CBSE Class 9 Science Exemplar Chapter 8: Motion NCERT Solutions

When a particle moves in a circular path of radius 'r' and completes half a circle, its initial and final positions are diametrically opposite. The displacement is defined as the shortest distance between the initial and final points. In this case, the displacement is equal to the diameter of the circle.

Distance travelled = $\backslash frac\{1\}\{2\}$ X circumference = $\backslash pi\; r$

Displacement = Diameter = 2r

We can use the third equation of motion: $v^2\; =\; u^2\; +\; 2as$.

Here, the final velocity 'v' at the greatest height is 0.

The acceleration 'a' is equal to '-g' because the body is moving against gravity.

The displacement 's' is the greatest height 'H'.

Substituting these values into the equation:

$$0\; =\; u^2\; +\; 2(-g)H$$

$$0\; =\; u^2\; -\; 2gH$$

Rearranging the equation to solve for H:

$$2gH\; =\; u^2$$

$$H\; =\; \backslash frac\{u^2\}\{2g\}$$

Displacement is the shortest distance between the initial and final points of a moving object, and it is a vector quantity. Distance is the total path length covered by the object, and it is a scalar quantity.

The magnitude of the displacement can never be greater than the distance covered. It can be equal to the distance only when the object moves in a straight line without changing its direction.

Therefore, the ratio of the magnitude of displacement to the distance is always less than or equal to 1.

Let the displacement 's' be proportional to the square of time 't'. This can be written as $s\; =\; kt^2$, where 'k' is a constant of proportionality.

The velocity 'v' is the rate of change of displacement with respect to time: $v\; =\; \backslash frac\{ds\}\{dt\}$.

$v\; =\; \backslash frac\{d\}\{dt\}(kt^2)\; =\; 2kt$

Since velocity $v\; =\; 2kt$, the velocity is directly proportional to time.

The acceleration 'a' is the rate of change of velocity with respect to time: $a\; =\; \backslash frac\{dv\}\{dt\}$.

$a\; =\; \backslash frac\{d\}\{dt\}(2kt)\; =\; 2k$

Since the acceleration 'a' is equal to a constant value (2k), the object moves with uniform acceleration.

The provided text mentions a 'v - t graph (Fig. 8.1)' but the figure itself is missing from the source. However, based on the options, we can infer the nature of the graph that would lead to each conclusion.

If the graph were a horizontal line, it would indicate uniform velocity (option a).

If the graph were a horizontal line at v=0, it would indicate the object is at rest (option b).

If the graph were a curve, it would indicate non-uniform motion or non-uniform acceleration (option c).

If the graph is a straight line passing through the origin or a straight line with a constant positive or negative slope (not horizontal), it represents uniform acceleration. The velocity changes at a constant rate.

Assuming the intended graph shows a straight line with a constant slope, the object is moving with uniform acceleration.