Motion in a Straight Line

A body can be treated as a point object when its size is much smaller than the distance it moves, and its shape/rotation do not matter. Among the given examples, a railway carriage moving smoothly between two stations can be approximated as a point object.

The other cases involve important effects of rotation, turning, or changing shape, so they cannot be treated as point objects.

From the x-t graph in Fig. 2.9:

- The child whose home is at the smaller distance from school lives closer to the school. From the graph, B is closer than A.
- The child whose graph starts earlier from the origin starts from school earlier. Here B starts earlier than A.
- The steeper graph means greater speed. A has the steeper slope, so A walks faster than B.
- The two children reach home at different times, as their graphs end at different times.
- The graphs intersect only once, so A overtakes B once on the road.

Given:
- Initial speed, $u = 126\ \text{km h}^{-1} = 126 \times \frac{5}{18} = 35\ \text{m s}^{-1}$
- Final speed, $v = 0$
- Distance, $s = 200\ \text{m}$

Using
$$v^2 = u^2 + 2as$$

$$0 = 35^2 + 2a(200)$$

$$0 = 1225 + 400a$$

$$a = -\frac{1225}{400} = -3.0625\ \text{m s}^{-2}$$

So the retardation has magnitude $3.06\ \text{m s}^{-2}$.

Now use
$$v = u + at$$

$$0 = 35 + (-3.0625)t$$

$$t = \frac{35}{3.0625} \approx 11.4\ \text{s}$$

So the stopping time is about $11.4\ \text{s}$.

Note: If using the textbook's own printed exercise, the standard uniform-retardation result is the computed value above.

For upward motion, the acceleration is always due to gravity, so it acts downward.

At the highest point:
- velocity = 0
- acceleration = g = 9.8\ \text{m s}^{-2} downward

Now take the highest point as origin and downward as positive.

### Upward motion
Since upward is opposite to the positive direction:
- position is negative
- velocity is negative
- acceleration is positive

### Downward motion
In the downward direction:
- position is positive
- velocity is positive
- acceleration is positive

### Height reached
Using
$$v^2 = v_0^2 + 2as$$

At the top, $v=0$, $v_0=29.4\ \text{m s}^{-1}$, $a=-9.8\ \text{m s}^{-2}$:
$$0 = (29.4)^2 + 2(-9.8)h$$
$$h = \frac{(29.4)^2}{2\times 9.8} = \frac{864.36}{19.6} = 44.1\ \text{m}$$

### Time to return
Time up =
$$t = \frac{v-v_0}{a} = \frac{0-29.4}{-9.8} = 3\ \text{s}$$

Total time back to the hands is twice this:
$$2\times 3 = 6\ \text{s}$$

- (a) True: A particle can be momentarily at rest, so its speed is zero, but it may still have non-zero acceleration. Example: a ball thrown upward has zero speed at the highest point, but acceleration due to gravity is non-zero.
- (b) False: If speed is zero, the magnitude of velocity is zero; hence velocity cannot be non-zero.
- (c) False: Constant speed does not necessarily mean zero acceleration. If direction changes, acceleration can be present. Example: uniform circular motion.
- (d) False: Positive acceleration does not always mean speeding up. If velocity is negative, positive acceleration may reduce the speed.

### (a) Magnitude of displacement and total path length
- Magnitude of displacement is the straight-line change in position between the initial and final points.
- Path length is the actual length of the path travelled.

For one-dimensional motion, the path length is always greater than or equal to the magnitude of displacement.
- Equality holds when the particle moves only in one direction without turning back.

### (b) Magnitude of average velocity and average speed
- Magnitude of average velocity =
$$\left|\frac{\text{displacement}}{\text{time}}\right|$$
- Average speed =
$$\frac{\text{total path length}}{\text{time}}$$

Since path length $\ge$ magnitude of displacement,
$$\text{average speed} \ge \text{magnitude of average velocity}$$

Equality holds when the motion is along a straight line in one direction, without reversal.