1 2 3 C

Motion I

7 (a) From 1 January 2009 to 10 January 2009,

the watch runs slower than the actual time by 9 minutes. Therefore, when the actual time is 2:00 pm on 10 January 2009, the time shown on the watch should be 1:51 pm on 10 January 2009.

Practice 1.1 (p. 6)

D (a) Possible percentage error 10 −6 = × 100% 24 × 3600 = 1.16 × 10 % 1 (b) = 1 000 000 days 10 −6

–9

It would take 1 000 000 days to be in error by 1 s.

(b) Percentage error 9 = × 100% 9 × 24 × 60

= 6.94 × 10–2%

4

(a) One day

= 24 × 60 × 60 = 86 400 s

Practice 1.2 (p. 15)

1 2 3 4 5

C B D D

(b) One year

= 365 × 86 400 = 31 500 000 s

5

Let t be the period of time recorded by a stop-watch. Percentage error =

0.4 × 100% ≤ 1% t

t ≥ 40 s

(a) Total distance she travels 2 × × 10 2 × × 20 2 × × 15 + + = 2 2 2 = 141 m

(b) Magnitude of total displacement

= 10 × 2 + 20 × 2 + 15 × 2 = 90 m Direction: east Her total displacement is 90 m east.

The minimum period of time is 40 s.

6

(a) Percentage error error due to reaction time = × 100% time measured 0.3 = × 100% 10 = 3%

6 7

His total displacement is 0. With the notation in the figure below.

(b) From (a), the percentage error of a short

time interval (e.g. 10 s) measured by a stop-watch is very large. Since the time intervals of 110-m hurdles are very short in the Olympic Games, stop-watches are not used to avoid large percentage errors. Since ZX = ZY = 1 m, α = β = 60°. Therefore, XY = ZX = ZY = 1 m The magnitude of the displacement of the ball is 1 m. ©

8

(a) The distance travelled by the ball will be

longer if it takes a curved path.

7

(a) Length of the path

= 0.8 × 120 = 96 m

(b) No matter which path the ball takes, its

displacement remains the same.

(b) Length of AB along the dotted line 96 = 30.6 m = (c)

Magnitude of Jack’s average velocity 30.6 × 2 = = 0.51 m s–1 120

Practice 1.3 (p. 23)

1

B Total time 5000 5000 = + = 9821 s 1.4 0.8 5000 + 5000 = 1.02 m s–1 Average speed = 9821

Practice 1.4 (p. 31)

1 2

C B Final speed = 1.5 × 1 – 0.2 × 1 = 1.3 m s–1

2

C Total time = 9821 + 10 × 60 =10 421 s 5000 + 5000 Average speed = = 0.96 m s–1 10 421

3

A By a =

3

D When the spacecraft had just finished 1 revolution, the spacecraft returned to its starting point. Therefore, its displacement was zero and its average velocity was also zero.

v −u , t

v = u + at 36 = + ( −1.5) × 2 3.6

= 7 m s–1 = 7 × 3.6 km h–1 = 25.2 km h–1 Its speed after 2 s is 25.2 km h–1.

4 5

D

(a) Average speed 100 = = 10.3 m s–1 9.69 (b) Yes. This is because the magnitude of the displacement is equal to the distance in this case.

4

B Take the direction of the original path as positive. Average acceleration of the ball −10 − 17 = 0.8 = –33.8 m s–2 The magnitude of the average acceleration of the ball is 33.8 m s–2. v −u By a = , t 100 −0 v − u 3.6 t= = = 4.27 s a 6.5

6

(a) Two cars move with the same speed, e.g.

50 km h–1, but in opposite directions.

(b) A man runs around a 400-m playground.

When we calculate his average speed, we can take 400 m as the distance and his average speed is non-zero. But since his displacement is zero (he returns to his starting point), his average velocity is zero.

5

The shortest time it takes is 4.27 s.

©

6

Time / s

–1

4

0 2 4 6 17 8 22

D Average speed 80 + 60 = 5

= 28 km h–1 Average velocity =

Speed / m s 2 7 12 v − u 22 − 2 a= = 2.5 m s–2 = t 8

The acceleration of the car is 2.5 m s–2.

7

(a) I will choose ‘towards the left’ as the

positive direction.

80 2 + 60 2 5

(b) 5

= 20 km h–1 C Total time 10 10 = + 2 3 = 8.33 s

v −u , t u = v − at = 9 − (−2) × 3 = 15 m s–1

–1

(c)

By a =

Average speed 20 = 8.33

= 2.4 m s–1

Her average speed for the whole trip is 2.4 m s–1.

The initial velocity of the skater is 15 m s .

8

(a) The object initially moves towards the

left and accelerates towards the left. It will speed up....