Chapter 4: Describing Motion Around Us (Physics)

He made two round trips of 250 m each way:

$$\text{Distance}=250+250+250+250=1000\text{m}$$

He finally returned home, so his final position is the same as the starting position:

$$\text{Displacement}=0\text{m}$$

Distance = 1000 m; displacement = 0 m.

The fourth floor is $4\times 3=12$ m above the ground floor, and the second floor is $2\times 3=6$ m above it.

(i) Up 12 m, then down from the 4th to the 2nd floor, 6 m: $12+6=18$ m.

(ii) The final position is 6 m above the starting point: displacement $=6$ m, upwards.

(i) 18 m (ii) 6 m vertically upwards.

Yes. Acceleration is the rate of change of velocity, and velocity has both magnitude (speed) and direction. The speedometer shows only speed. If she goes round a curve or a roundabout, or turns a corner, at constant speed, the direction of motion changes, so the velocity changes and the scooter is accelerating (this is accelerated motion even though the speed is constant).

Yes. If she moves along a curved path at constant speed, her direction (and so her velocity) keeps changing, which means she is accelerating.

$u=0$, $v=24$ m s⁻¹, $t=6$ s.

$$a=\frac{v-u}{t}=\frac{24-0}{6}=4{\text{m s}}^{-2}$$

$$s=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}\times 4\times 36=72\text{m}$$

Acceleration = 4 m s⁻² (in the direction of motion); distance = 72 m.

$u=28$ m s⁻¹, $v=0$, $s=98$ m.

$$v^2=u^2+2as\Rightarrow 0=784+2a(98)\Rightarrow a=-\frac{784}{196}=-4{\text{m s}}^{-2}$$

$$t=\frac{v-u}{a}=\frac{0-28}{-4}=7\text{s}$$

Acceleration = −4 m s⁻² (opposite to the motion, i.e. retardation of 4 m s⁻²); time = 7 s.

No. In a position–time graph, the slope of the line gives the velocity. Both graphs are straight lines, so each object moves with a constant velocity, but line A is steeper than line B throughout. So A always moves faster than B. At t = 5 s the lines cross: A and B are at the same position at that instant (A overtakes B), but their velocities are still different.

No: both move with constant but different velocities (A's line is always steeper); at t = 5 s they only have the same position, not the same velocity.

Answer: (i) and (ii).

Both start at the same position at t = 0 and are at the same position at t = 10 s, so they have the same displacement in the same time: equal average velocity (i). Both keep moving forward (the positions only increase), so for each the distance equals the displacement: both cover equal distances in 10 s, so their average speeds are equal (ii). B is slower in some parts and faster in others, but this does not change the average.

(i) and (ii).

$u=54\times \frac{5}{18}=15$ m s⁻¹, $v=36\times \frac{5}{18}=10$ m s⁻¹, $t=36$ s. For constant acceleration:

$$s=\frac{u+v}{2}\times t=\frac{15+10}{2}\times 36=12.5\times 36=450\text{m}$$

(The retardation is $\frac{15-10}{36}\approx 0.14$ m s⁻².)

450 m

Use $s=\frac{u+v}{2}t$ for each stage (or the areas under the velocity–time graph):

• Accelerating: $\frac{0+20}{2}\times 5=50$ m
• Constant speed: $20\times 10=200$ m
• Braking: $\frac{20+0}{2}\times 6=60$ m

$$\text{Total}=50+200+60=310\text{m}$$

310 m

$u=36\times \frac{5}{18}=10$ m s⁻¹.

• During the reaction time the bus keeps moving at 10 m s⁻¹: $10\times 0.5=5$ m.
• While braking, with $a=-2.5$ m s⁻² and $v=0$: $s=\frac{v^2-u^2}{2a}=\frac{0-100}{-5}=20$ m.

$$\text{Stopping distance}=5+20=25\text{m}<30\text{m}$$

Yes. It stops in 25 m (5 m during reaction + 20 m braking), 5 m before the obstacle.

Rest and motion are relative: they depend on the reference point (frame of reference).

• With respect to the Earth (the room, the ground), the object does not change its position, so it is at rest.
• With respect to the Sun, the object moves along with the Earth (around the Sun at about 30 km s⁻¹, and also with the Earth's rotation), so it is in motion.

So an object on the Earth can be considered at rest only when we choose the Earth as the reference; no object is at rest in an absolute sense.

Yes, relative to the Earth it is at rest; relative to the Sun it is moving with the Earth. Rest and motion depend on the chosen reference point.

From Fig. 4.30: the velocity rises from 0 to 3 m s⁻¹ in the first 20 s, stays at 3 m s⁻¹ from 20 s to 100 s, and falls to 2 m s⁻¹ from 100 s to 120 s.

Displacement = total area under the graph:

$$\frac{1}{2}\times 20\times 3+80\times 3+\frac{3+2}{2}\times 20=30+240+50=320\text{m}$$

(The shaded areas are (i) 240 m and (ii) 50 m.)

Average acceleration over 120 s:

$$a=\frac{v_{\text{final}}-v_{\text{initial}}}{t}=\frac{2-0}{120}=\frac{1}{60}\approx 0.017{\text{m s}}^{-2}$$

Displacement = 320 m (240 m at constant velocity, 50 m while slowing down); average acceleration = 1/60 m s⁻² ≈ 0.017 m s⁻².

The distance is the area under the velocity–time graph. Reading the graph approximately and splitting it into strips:

$$\text{Distance}\approx 3.5+7.25+11.25+10.9+6.75+6.5\approx 46\text{km}$$

(A quick check: average speed about 7 km h⁻¹ for about 6.5 h gives about 45 km. Your estimate may differ slightly depending on how you read the graph.)

About 45–46 km (area under the velocity–time graph).

In the last 6 s the velocity rises from 6 m s⁻¹ to $6+1\times 6=12$ m s⁻¹.

Area under the graph:

$$\text{Rectangle: }6\times 120=720\text{m},\text{Trapezium: }\frac{6+12}{2}\times 6=54\text{m}$$

$$\text{Displacement}=720+54=774\text{m}$$

774 m

Accelerations: $a_A=\frac{5}{5}=1$ m s⁻², $a_B=\frac{3}{10}=0.3$ m s⁻².

Displacement = area of each triangle:

$$A:\frac{1}{2}\times 5\times 5=12.5\text{m},B:\frac{1}{2}\times 10\times 3=15\text{m}$$

Car A: 12.5 m in 5 s; car B: 15 m in 10 s.

In 1 h 30 min the minute hand makes $1\frac{1}{2}$ rounds. At 6:00 it points to 12; at 7:30 it points to 6, diametrically opposite.

(i) $$\text{Distance}=1.5\times 2\pi r=1.5\times 2\times \frac{22}{7}\times 7=66\text{cm}$$

(ii) Displacement = straight-line distance from "12" to "6" = diameter $=2\times 7=14$ cm (from 12 towards 6).

(iii) Time $=1.5\text{h}=5400$ s.

$$\text{Average speed}=\frac{66}{5400}=\frac{11}{900}\approx 0.012{\text{cm s}}^{-1}$$

(iv) $$\text{Average velocity}=\frac{14}{5400}=\frac{7}{2700}\approx 0.0026{\text{cm s}}^{-1}\text{(from 12 towards 6)}$$

(i) 66 cm (ii) 14 cm (iii) 11/900 cm s⁻¹ ≈ 0.012 cm s⁻¹ (iv) 7/2700 cm s⁻¹ ≈ 0.0026 cm s⁻¹, directed from 12 towards 6.