Transcript M2.2 Kinematics - Haringeymath's Blog

AS-Level Maths:
Mechanics 2
for Edexcel
M2.2 Kinematics
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Variable acceleration
Contents
Variable acceleration
Motion in two or three dimensions
Examination-style questions
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Variable acceleration
Kinematics is the study of how objects move rather
than of the forces that cause motion.
In M1 most situations involved constant acceleration.
However, the acceleration of a particle is not always constant.
In cases of variable acceleration the displacement,
velocity and acceleration will be functions of time.
Calculus must be used to change between displacement,
velocity and acceleration.
x = f (t)
v = f ’(t)
a = f ’’(t)
where x is displacement, v is velocity and a is acceleration.
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Variable acceleration
To find a where acceleration is variable, differentiate v
with respect to t.
To find x, integrate v with respect to t.
To find v, either differentiate x or integrate a with respect
to t.
When integrating, remember to include the constant, c. If
we know the initial velocity or position of the particle we
can use this information to find c.
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Variable acceleration
The difference between constant and variable acceleration
may not be discernible to the eye. Which ball is travelling
with variable acceleration?
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Variable acceleration question 1
Question 1: The displacement of a particle moving in a
straight line is given by x = t3 – 3t2 + 4t.
Find
a) the velocity of the particle when t = 4.
b) the acceleration of the particle when t = 2.
a) To find the velocity we must differentiate x:
dx
v=
dt
v = 3t2 – 6t + 4
When t = 4: v = 3 × 42 – 6 × 4 + 4 = 28
Therefore the velocity of the particle when t = 4 is 28 ms-1.
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Variable acceleration question 1
b) To find the acceleration we must first differentiate v:
v = 3t2 – 6t + 4
dv
a=
dt
a = 6t – 6
When t = 2: a = 6 × 2 – 6 = 6
Therefore the acceleration of the particle when t = 2 is 6 ms-2.
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Variable acceleration question 2
Question 2: The velocity of a particle is given by v = t2 – 5t + 6.
Initially its displacement from O is 2 m.
Find
a) the acceleration of the particle after 4 seconds.
b) the displacement of the particle from O when t = 6.
a) Differentiate v:
dv
a=
dt
a = 2t – 5
When t = 4: a = 2 × 4 – 5
a=3
Therefore the acceleration of the particle when t = 4 is 3 ms-2.
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Variable acceleration question 2
b) To find the displacement we need to integrate v:
3
2
t
5
t

 6t  c
x    t 2  5t  6 dt 
3
2
The initial displacement from O is 2 m, i.e. when t = 0, x = 2:
2=0–0+0+c
c=2
t 3 5t 2

 6t  2
 x
3
2
When t = 6: x = 72 – 90 + 36 + 2
= 20
Therefore when t = 6 the displacement of the
particle from O is 20 m.
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Variable acceleration question 3
Question 3: The velocity of a particle travelling in a straight
line is given by v = –3t2 + 12t + 4. Find:
a) the time when the particle is travelling at its maximum
velocity.
b) the value of this maximum velocity.
a) Here v is a quadratic where the coefficient for t2 is negative.
Therefore the maximum point on the velocity curve occurs
where the gradient is 0.
Differentiate to find the gradient function:
dv
= –6t + 12
dt
–6t + 12 = 0
t=2
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Variable acceleration question 3
Question 3: The velocity of a particle travelling in a straight
line is given by v = –3t2 + 12t + 4. Find:
a) the time when the particle is travelling at its maximum
velocity.
b) the value of this maximum velocity.
b) When t = 2:
v = –3 × 22 + 12 × 2 + 4
v = 16
Therefore the maximum velocity of the particle occurs
when t = 2 seconds and the maximum velocity is 16 ms-1.
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Motion in two or three dimensions
Contents
Variable acceleration
Motion in two or three dimensions
Examination-style questions
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Motion in two or three dimensions
The motion discussed so far has taken place in one dimension.
However, motion can also occur in two or three dimensions.
In these cases displacement, velocity and acceleration are
represented by the unit vectors i, j and k.
r = xi + yj + zk
v = x`i + y`j + z`k
a = x``i + y``j + z``k
where x, y and z are all functions of time.
When integrating v or a as unit vectors, we need to
remember the constants c, d and e, one for each dimension.
i.e. ci + dj + ek.
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3-D motion
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Vector question 1
Question 1: The position vector of a particle at time t is given
by
r = (2t – 4)i + (t2 – 3t + 1)j
a) Find the velocity and acceleration of the particle at time t.
b) Find the velocity and acceleration of the particle when t = 4.
a) Differentiate to find v:
Differentiate to find a:
b) When t = 4:
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dr
v=
dt
v = 2i + (2t – 3)j
dv
a=
dt
v = 2i + 5j ms-1
a = 2j
a = 2j ms-2
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Vector question 2
Question 2: The velocity of a particle at time t is given by
v = 3t2i + (2t – 7)j
Find the position vector and acceleration of the particle at time
t if the position vector of the particle at time t = 0 is 3i – 2j.
Differentiate to find the acceleration:
dv
a=
dt
a = 6ti + 2j
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Vector question 2
Question 2: The velocity of a particle at time t is given by
v = 3t2i + (2t – 7)j
Find the position vector and acceleration of the particle at time
t if the position vector of the particle at time t = 0 is 3i – 2j.
Integrate to find the position vector:
r    3t 2i  (2t  7) jdt
r = (t3 + c)i + (t2 – 7t + d)j
When t = 0: r = 3i – 2j
ci + dj = 3i – 2j
 c = 3 and d = –2
r = (t3 + 3)i + (t2 – 7t – 2)j
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Vector question 3
Question 3: The position vector of a particle is given by
r = 2t2i + 4tj + 3k
Find the speed of the particle when t = 3.
Differentiate to find the velocity:
v
12
dr
v=
dt
v = 4ti + 4j
When t = 3: v = 12i + 4j
Speed = (122 + 42)
4
= 160
= 12.6 (3 s.f.)
Therefore the speed of the particle when t = 3 is 12.6 ms-1.
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Examination-style questions
Contents
Variable acceleration
Motion in two or three dimensions
Examination-style questions
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Exam question 1
Question 1: A particle has position vector at time t seconds
given by
r = (2t2 – 3t)i + (t3 – t2 + 4)j
a) Find the velocity of the particle at time t.
b) Find the acceleration of the particle at time t = 3.
a) Differentiate to find v:
dr
v=
dt
v = (4t – 3)i + (3t2 – 2t)j
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Exam question 1
Question 1: A particle has position vector at time t seconds
given by
r = (2t2 – 3t)i + (t3 – t2 + 4)j
a) Find the velocity of the particle at time t.
b) Find the acceleration of the particle at time t = 3.
b) Differentiate to find a:
dv
a=
dt
a = 4i + (6t – 2)j
When t = 3: a = 4i + 16j
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Exam question 2
Question 2: A particle is moving with acceleration a = 3ti – 2j
at time t seconds.
Given that at time t = 0 the particle is at 4i + 2j and has velocity
i – 4j, find:
a) the velocity of the particle at time t.
b) the position vector of the particle at time t.
c) the distance of the particle from the origin when t = 3.
a) Integrate to find v:
v    3ti  2 jdt
3t 2
v
i  2t j  ci  d j
2
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Exam question 2
When t = 0: v = i – 4j
0 + 0 + ci + dj = i – 4j  c = 1 and d = –4
 3t 2 
 v
 1 i   2t  4  j
 2



 3t 2


r   
 1 i   2t  4  jdt
b) Integrate to find r:

 2

 t3 
r    t  i  t 2  4t j  ci  d j
2



When t = 0: r = 4i + 2j
4i + 2j = ci + dj
 c = 4 and d = 2
 t3

 r    t  4  i  t 2  4t  2 j
2


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
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Exam question 2
Question 2: A particle is moving with acceleration a = 3ti – 2j
at time t seconds.
Given that at time t = 0 the particle is at 4i + 2j and has velocity
i – 4j, find
a) the velocity of the particle at time t.
b) the position vector of the particle at time t.
c) the distance of the particle from the origin when t = 3.
 27

 3  4  i   9  12  2  j = 20.5i – 19j
c) When t = 3: r = 
 2

 The distance from the origin = (20.52 + 192)
= 28.0 m (3 s.f.)
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Exam question 3
Question 3: The position vector of a particle at time t is given
by
r = (sin t)i + (cos t)j + k
0≤t≤

a) Find the speed of the particle when t = .
4
b) Find the time when the velocity is parallel to j.
dr
v=
dt
a) Differentiate to find v:
v = (cos t)i – (sin t)j
When t =
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
4
:
v=
1
1
i
j
2
2
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Exam question 3
2
Speed =v=
 1   1 

 

 2  2
2
v = (½ + ½) = 1
Therefore the speed when t =

4
is 1 ms-1.
b) v is parallel to j when the i component of v is 0.
v = (cos t)i – (sin t)j
 cos t = 0

t=
2

 v is parallel to j when t =
seconds.
2
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