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Physics Form 4 Chapter 2 Force and Motion I

(–10.0 – 0) SPM Highlights
Gradient =
(25.0 – 20.0)
= –20.0 m s –2 The v-t graph in Figure 2.32 shows the situation of
a skydiver when jumping from a plane until he lands
Note: In this case, even though the velocity on the ground. Analyse the v-t graph and explain the
of the motorcycle has an increasing motion of the skydiver when
magnitude, its acceleration is negative. (a) the parachute is not open,
(b) the parachute is open,
This is because the negative indicates (c) the skydiver lands,
the motorcycle moves in the opposite v / m s –1
direction.
50
(b) (i) The displacement of the motorcycle after 40
20.0 s = area under the graph for t = 0 s to Parachute Parachute
t = 20.0 s 30 not opened opens Chapter
1
s =  × 10.0 × 25.0 + (5.0 × 25.0) + 20 Landing
2
1
 × 5.0 × 25.0 10 0 t /s 2
2
= 312.5 m 5 10 15 20 25 30 35 40
Figure 2.32
t = 0 s
t = 20.0 s Answer:
s = 312.5 m
t = 25.0 s (a) When the parachute is not open, the skydiver
(ii) For t = 20.0 s to t = 25.0 s, the motorcycle falls with increasing velocity until it reaches
moves in the opposite direction as shown a maximum velocity which is known as the
terminal velocity.
in the diagram. The distance travelled by (b) When the parachute is open, the skydiver fall
the motorcycle for t = 20.0 s to 25.0 s with a uniform deceleration until it reaches
1 minimum velocity upon which he continues to
s = (5.0 × 10.0) fall with this velocity.
2
= 25.0 m (c) When the skydiver lands, stop falling and his
velocity descreases until 0 m s .
–1
Therefore, the displacement of the
motorcycle after 25.0 s
s = 312.5 – 25.0
= 287.5 m Checkpoint 2.2
Q1 The s-t graph in Figure 2.33 shows the movement
(c)
a / m s –2 of a crate that is being hoisted vertically by a
crane. s is the displacement of the crate from the
3.0 ground.
2.0
1.0 s / m
t /s
0 2.0
–1.0 5.0 10.0 15.0 20.0 25.0
–2.0 1.5
–3.0
1.0
–4.0
s
–5.0 0.5
0 1 2 3 4 5 t / s
SPM Tips Figure 2.33
Make sure you can use the acceleration formula, (a) Briefly describe the movement of the crate
v – u from t = 0 to 5 s.
a = to calculate acceleration from
t (b) Calculate the velocity of the crate from t = 2.4
• ticker tape, to 3.8 s.
• ticker chart, and (c) What is the average velocity of the crate
• velocity-time graph. from t = 0 to 5 s?

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