12 x1 t07 01 v and a in terms of x
TRANSCRIPT
Velocity & Acceleration in Terms of x
Velocity & Acceleration in Terms of x
If v = f(x);
Velocity & Acceleration in Terms of x
If v = f(x);
= 2
2
2
2
1v
dx
d
dt
xd
Velocity & Acceleration in Terms of x
If v = f(x);
= 2
2
2
2
1v
dx
d
dt
xd
Proof:
dtdv
dtxd =2
2
Velocity & Acceleration in Terms of x
If v = f(x);
= 2
2
2
2
1v
dx
d
dt
xd
Proof:
dtdv
dtxd =2
2
dt
dx
dx
dv ⋅=
Velocity & Acceleration in Terms of x
If v = f(x);
= 2
2
2
2
1v
dx
d
dt
xd
Proof:
dtdv
dtxd =2
2
dt
dx
dx
dv ⋅=
vdx
dv ⋅=
Velocity & Acceleration in Terms of x
If v = f(x);
= 2
2
2
2
1v
dx
d
dt
xd
Proof:
dtdv
dtxd =2
2
dt
dx
dx
dv ⋅=
vdx
dv ⋅=
⋅= 2
21
vdvd
dxdv
Velocity & Acceleration in Terms of x
If v = f(x);
= 2
2
2
2
1v
dx
d
dt
xd
Proof:
dtdv
dtxd =2
2
dt
dx
dx
dv ⋅=
vdx
dv ⋅=
⋅= 2
21
vdvd
dxdv
= 2
21
vdxd
e.g. (i) A particle moves in a straight line so thatFind its velocity in terms of x given that v = 2 when x = 1.
xx 23−=
e.g. (i) A particle moves in a straight line so thatFind its velocity in terms of x given that v = 2 when x = 1.
xvdxd
2321 2 −=
xx 23−=
e.g. (i) A particle moves in a straight line so thatFind its velocity in terms of x given that v = 2 when x = 1.
xx 23−=
xvdxd
2321 2 −=
cxxv +−= 22 321
e.g. (i) A particle moves in a straight line so thatFind its velocity in terms of x given that v = 2 when x = 1.
xx 23−=
xvdxd
2321 2 −=
cxxv +−= 22 321
( ) ( )
0
113221
i.e.
2,1when
22
=
+−=
==
c
c
vx
e.g. (i) A particle moves in a straight line so thatFind its velocity in terms of x given that v = 2 when x = 1.
xx 23−=
xvdxd
2321 2 −=
cxxv +−= 22 321
( ) ( )
0
113221
i.e.
2,1when
22
=
+−=
==
c
c
vx
22 26 xxv −=∴
e.g. (i) A particle moves in a straight line so thatFind its velocity in terms of x given that v = 2 when x = 1.
xx 23−=
xvdxd
2321 2 −=
cxxv +−= 22 321
( ) ( )
0
113221
i.e.
2,1when
22
=
+−=
==
c
c
vx
22 26 xxv −=∴
226 xxv −±=
e.g. (i) A particle moves in a straight line so thatFind its velocity in terms of x given that v = 2 when x = 1.
xx 23−=
xvdxd
2321 2 −=
cxxv +−= 22 321
( ) ( )
0
113221
i.e.
2,1when
22
=
+−=
==
c
c
vx
22 26 xxv −=∴
226 xxv −±=
NOTE:
02 ≥v
e.g. (i) A particle moves in a straight line so thatFind its velocity in terms of x given that v = 2 when x = 1.
xx 23−=
xvdxd
2321 2 −=
cxxv +−= 22 321
( ) ( )
0
113221
i.e.
2,1when
22
=
+−=
==
c
c
vx
22 26 xxv −=∴
226 xxv −±=
NOTE:
02 ≥v
026 2 ≥− xx
e.g. (i) A particle moves in a straight line so thatFind its velocity in terms of x given that v = 2 when x = 1.
xx 23−=
xvdxd
2321 2 −=
cxxv +−= 22 321
( ) ( )
0
113221
i.e.
2,1when
22
=
+−=
==
c
c
vx
22 26 xxv −=∴
226 xxv −±=
NOTE:
02 ≥v
026 2 ≥− xx
( ) 032 ≥− xx
e.g. (i) A particle moves in a straight line so thatFind its velocity in terms of x given that v = 2 when x = 1.
xx 23−=
xvdxd
2321 2 −=
cxxv +−= 22 321
( ) ( )
0
113221
i.e.
2,1when
22
=
+−=
==
c
c
vx
22 26 xxv −=∴
226 xxv −±=
NOTE:
02 ≥v
026 2 ≥− xx
( ) 032 ≥− xx
30 ≤≤ x
e.g. (i) A particle moves in a straight line so thatFind its velocity in terms of x given that v = 2 when x = 1.
xx 23−=
xvdxd
2321 2 −=
cxxv +−= 22 321
( ) ( )
0
113221
i.e.
2,1when
22
=
+−=
==
c
c
vx
22 26 xxv −=∴
226 xxv −±=
NOTE:
02 ≥v
026 2 ≥− xx
( ) 032 ≥− xx
30 ≤≤ x
∴Particle moves between x = 0 and x = 3 and nowhere else.
23xx =m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is 1 unit to the right of O, and is traveling with a velocity of in the negative direction. Find x in terms of t.
23xx =m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is 1 unit to the right of O, and is traveling with a velocity of in the negative direction. Find x in terms of t.
22 321
xvdxd =
23xx =m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is 1 unit to the right of O, and is traveling with a velocity of in the negative direction. Find x in terms of t.
22 321
xvdxd =
cxv += 32
21
23xx =m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is 1 unit to the right of O, and is traveling with a velocity of in the negative direction. Find x in terms of t.
22 321
xvdxd =
( )0
1221
i.e.
2,1,0when
32
=
+=−
−===
c
c
vxt
cxv += 32
21
23xx =m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is 1 unit to the right of O, and is traveling with a velocity of in the negative direction. Find x in terms of t.
22 321
xvdxd =
( )0
1221
i.e.
2,1,0when
32
=
+=−
−===
c
c
vxt
cxv += 32
21
3
32
2
2
xv
xv
±=
=∴
23xx =m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is 1 unit to the right of O, and is traveling with a velocity of in the negative direction. Find x in terms of t.
22 321
xvdxd =
( )0
1221
i.e.
2,1,0when
32
=
+=−
−===
c
c
vxt
cxv += 32
21
3
32
2
2
xv
xv
±=
=∴
32xdtdx −=
23xx =m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is 1 unit to the right of O, and is traveling with a velocity of in the negative direction. Find x in terms of t.
22 321
xvdxd =
( )0
1221
i.e.
2,1,0when
32
=
+=−
−===
c
c
vxt
cxv += 32
21
3
32
2
2
xv
xv
±=
=∴
32xdtdx −=
(Choose –ve to satisfy the initial conditions)
23xx =m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is 1 unit to the right of O, and is traveling with a velocity of in the negative direction. Find x in terms of t.
22 321
xvdxd =
( )0
1221
i.e.
2,1,0when
32
=
+=−
−===
c
c
vxt
cxv += 32
21
3
32
2
2
xv
xv
±=
=∴
32xdtdx −=
2
3
2x−=
(Choose –ve to satisfy the initial conditions)
23xx =m/s2
(ii) A particle’s acceleration is given by . Initially, the particle is 1 unit to the right of O, and is traveling with a velocity of in the negative direction. Find x in terms of t.
22 321
xvdxd =
( )0
1221
i.e.
2,1,0when
32
=
+=−
−===
c
c
vxt
cxv += 32
21
3
32
2
2
xv
xv
±=
=∴
32xdtdx −=
2
3
2x−=
(Choose –ve to satisfy the initial conditions)
2
3
21 −
−= xdxdt
2
3
21 −
−= xdxdt
2
3
21 −
−= xdxdt
cx
cx
cxt
+=
+=
+−⋅−=
−
−
2
2
22
1
2
1
2
1
2
3
21 −
−= xdxdt
cx
cx
cxt
+=
+=
+−⋅−=
−
−
2
2
22
1
2
1
2
1
when t = 0, x = 1
2
3
21 −
−= xdxdt
cx
cx
cxt
+=
+=
+−⋅−=
−
−
2
2
22
1
2
1
2
1
when t = 0, x = 1
2
20 i.e.
−=+=
c
c
2
3
21 −
−= xdxdt
cx
cx
cxt
+=
+=
+−⋅−=
−
−
2
2
22
1
2
1
2
1
when t = 0, x = 1
2
20 i.e.
−=+=
c
c
22 −=x
t
2
3
21 −
−= xdxdt
cx
cx
cxt
+=
+=
+−⋅−=
−
−
2
2
22
1
2
1
2
1
when t = 0, x = 1
2
20 i.e.
−=+=
c
c
22 −=x
t
OR
2
3
21 −
−= xdxdt
cx
cx
cxt
+=
+=
+−⋅−=
−
−
2
2
22
1
2
1
2
1
when t = 0, x = 1
2
20 i.e.
−=+=
c
c
22 −=x
t
OR ∫−
−=x
dxxt1
2
3
21
2
3
21 −
−= xdxdt
cx
cx
cxt
+=
+=
+−⋅−=
−
−
2
2
22
1
2
1
2
1
when t = 0, x = 1
2
20 i.e.
−=+=
c
c
22 −=x
t
OR ∫−
−=x
dxxt1
2
3
21
x
x1
2
1
22
1
−−=
−
2
3
21 −
−= xdxdt
cx
cx
cxt
+=
+=
+−⋅−=
−
−
2
2
22
1
2
1
2
1
when t = 0, x = 1
2
20 i.e.
−=+=
c
c
22 −=x
t
OR ∫−
−=x
dxxt1
2
3
21
x
x1
2
1
22
1
−−=
−
−= 1
12
x
2
3
21 −
−= xdxdt
cx
cx
cxt
+=
+=
+−⋅−=
−
−
2
2
22
1
2
1
2
1
when t = 0, x = 1
2
20 i.e.
−=+=
c
c
22 −=x
t
OR ∫−
−=x
dxxt1
2
3
21
x
x1
2
1
22
1
−−=
−
−= 1
12
x
xt
22 =+
2
3
21 −
−= xdxdt
cx
cx
cxt
+=
+=
+−⋅−=
−
−
2
2
22
1
2
1
2
1
when t = 0, x = 1
2
20 i.e.
−=+=
c
c
22 −=x
t
OR ∫−
−=x
dxxt1
2
3
21
x
x1
2
1
22
1
−−=
−
−= 1
12
x
xt
22 =+
( ) 22
2 += tx
2
3
21 −
−= xdxdt
cx
cx
cxt
+=
+=
+−⋅−=
−
−
2
2
22
1
2
1
2
1
when t = 0, x = 1
2
20 i.e.
−=+=
c
c
22 −=x
t
OR ∫−
−=x
dxxt1
2
3
21
x
x1
2
1
22
1
−−=
−
−= 1
12
x
xt
22 =+
( ) 22
2 += tx
( ) 22
2
+=
tx
A particle is moving along the x axis starting from a position 2 metres to the right of the origin (that is, x = 2 when t = 0) with an initial velocity of 5 m/s and an acceleration given by
2004 Extension 1 HSC Q5a)
xxx 22 3 +=(i) Show that 12 += xx
A particle is moving along the x axis starting from a position 2 metres to the right of the origin (that is, x = 2 when t = 0) with an initial velocity of 5 m/s and an acceleration given by
2004 Extension 1 HSC Q5a)
xxx 22 3 +=(i) Show that 12 += xx
xxvdxd
2221 32 +=
A particle is moving along the x axis starting from a position 2 metres to the right of the origin (that is, x = 2 when t = 0) with an initial velocity of 5 m/s and an acceleration given by
2004 Extension 1 HSC Q5a)
xxx 22 3 +=(i) Show that 12 += xx
xxvdxd
2221 32 +=
cxxv ++= 242
21
21
A particle is moving along the x axis starting from a position 2 metres to the right of the origin (that is, x = 2 when t = 0) with an initial velocity of 5 m/s and an acceleration given by
2004 Extension 1 HSC Q5a)
xxx 22 3 +=(i) Show that 12 += xx
xxvdxd
2221 32 +=
cxxv ++= 242
21
21
When x = 2, v = 5
A particle is moving along the x axis starting from a position 2 metres to the right of the origin (that is, x = 2 when t = 0) with an initial velocity of 5 m/s and an acceleration given by
2004 Extension 1 HSC Q5a)
xxx 22 3 +=(i) Show that 12 += xx
xxvdxd
2221 32 +=
cxxv ++= 242
21
21
When x = 2, v = 5
( ) ( ) ( )1 125 16 4
2 21
2
c
c
= + +
=
A particle is moving along the x axis starting from a position 2 metres to the right of the origin (that is, x = 2 when t = 0) with an initial velocity of 5 m/s and an acceleration given by
2004 Extension 1 HSC Q5a)
xxx 22 3 +=(i) Show that 12 += xx
xxvdxd
2221 32 +=
cxxv ++= 242
21
21
When x = 2, v = 5
( ) ( ) ( )1 125 16 4
2 21
2
c
c
= + +
=
12 242 ++= xxv
A particle is moving along the x axis starting from a position 2 metres to the right of the origin (that is, x = 2 when t = 0) with an initial velocity of 5 m/s and an acceleration given by
2004 Extension 1 HSC Q5a)
xxx 22 3 +=(i) Show that 12 += xx
xxvdxd
2221 32 +=
cxxv ++= 242
21
21
When x = 2, v = 5
( ) ( ) ( )1 125 16 4
2 21
2
c
c
= + +
=
12 242 ++= xxv
( )222 1+= xv
A particle is moving along the x axis starting from a position 2 metres to the right of the origin (that is, x = 2 when t = 0) with an initial velocity of 5 m/s and an acceleration given by
2004 Extension 1 HSC Q5a)
xxx 22 3 +=(i) Show that 12 += xx
xxvdxd
2221 32 +=
cxxv ++= 242
21
21
When x = 2, v = 5
( ) ( ) ( )1 125 16 4
2 21
2
c
c
= + +
=
12 242 ++= xxv
( )222 1+= xv
12 += xv
A particle is moving along the x axis starting from a position 2 metres to the right of the origin (that is, x = 2 when t = 0) with an initial velocity of 5 m/s and an acceleration given by
2004 Extension 1 HSC Q5a)
xxx 22 3 +=(i) Show that 12 += xx
xxvdxd
2221 32 +=
cxxv ++= 242
21
21
When x = 2, v = 5
( ) ( ) ( )1 125 16 4
2 21
2
c
c
= + +
=
12 242 ++= xxv
( )222 1+= xv
12 += xv
Note: v > 0, in order to satisfy initial
conditions
(ii) Hence find an expression for x in terms of t
(ii) Hence find an expression for x in terms of t
12 += xdtdx
(ii) Hence find an expression for x in terms of t
12 += xdtdx
∫∫ +=
xt
xdx
dt2
20 1
(ii) Hence find an expression for x in terms of t
12 += xdtdx
∫∫ +=
xt
xdx
dt2
20 1
[ ] xxt 2
1tan−=
(ii) Hence find an expression for x in terms of t
12 += xdtdx
∫∫ +=
xt
xdx
dt2
20 1
[ ] xxt 2
1tan−=
2tantan 11 −− −= xt
(ii) Hence find an expression for x in terms of t
12 += xdtdx
∫∫ +=
xt
xdx
dt2
20 1
[ ] xxt 2
1tan−=
2tantan 11 −− −= xt
2tantan 11 −− += tx
(ii) Hence find an expression for x in terms of t
12 += xdtdx
∫∫ +=
xt
xdx
dt2
20 1
[ ] xxt 2
1tan−=
2tantan 11 −− −= xt
2tantan 11 −− += tx
( )2tantan 1−+= tx
(ii) Hence find an expression for x in terms of t
12 += xdtdx
∫∫ +=
xt
xdx
dt2
20 1
[ ] xxt 2
1tan−=
2tantan 11 −− −= xt
2tantan 11 −− += tx
( )2tantan 1−+= tx
tt
xtan21
2tan−
+=
(ii) Hence find an expression for x in terms of t
12 += xdtdx
∫∫ +=
xt
xdx
dt2
20 1
[ ] xxt 2
1tan−=
2tantan 11 −− −= xt
2tantan 11 −− += tx
( )2tantan 1−+= tx
tt
xtan21
2tan−
+=
Exercise 3E; 1 to 3 acfh,7 , 9, 11, 13, 15, 17, 18,
20, 21, 24*