Page 5 - Pra U STPM 2021 Penggal 1 - Physics

Physics Term 1 STPM Chapter 2 Kinematics
2. For a body travelling along a straight line, if u = initial velocity, v = final Exam Tips
velocity after a time t, and s = displacement in time t, There are five quantities,
Change in velocity
then uniform acceleration, a = u, v, t, s, a, in the four
Time taken equations for motion
v – u under constant
= acceleration. Each of the
t
equations involves only
v = u + at .............................. a four of the quantities.
2 In a question, usually
Displacement, s = Average velocity × Time values for three of the
quantities are provided,
( ) and you are asked to
u + v
s = 2 t .............................. b calculate the value of
another of the quantities.
( u + u + at Use the equation
2 )
s = t that contains the four
quantities mentioned in
the question.
2 2
1
s = ut + at .............................. c
Info Physics
2s
From ①, v – u = at From ➁, v + u = t A380 Airbus take-off speed
(v – u)(v + u) = at × 2s A wide-body aircraft such as
t the Airbus A380 has a take-
–1
off speed of 280 km h and
v – u = 2as .............................. d ➃ requires a runway of length
2
2
3000 m. What then is its ac-
celeration?
Example 1

–1
A motorist travelling at 72 km h on a straight road approaches a traffic light which turns red
when he is 55.0 m away from the stop line. The reaction time is 0.7 s. With the brakes applied
fully, the vehicle decelerates at 5.0 m s . How far from the stop line will he stop and on which
–2
side of the stop line?
Solution:

–1
72 km h = 72 000 m = 20 m s –1
3 600 s
–1
The vehicle travels at constant speed of 20 m s during the reaction time of 0.7 s.
Distance travelled during the reaction time = 20 × 0.70 = 14.0 m
When the vehicle stops, final velocity v = 0
Acceleration = –5.0 m s (negative because of deceleration)
–2
Using v = u + 2as
2
2
2
Distance travelled when the brakes are applied, s = v – u 2
2a
= 0 – 20 2
2 (–5)
= 40.0 m
Total distance travelled = (14.0 + 40.0) m = 54.0 m
Hence, the vehicle stops (55.0 – 54.0) m = 1.0 m before the stop line.

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