12 x1 t07 02 v and a in terms of x (2012)
TRANSCRIPT
Velocity & Acceleration in Terms of x
If v = f(x);
2
2
2
21 v
dxd
dtxd
Proof:
dtdv
dtxd2
2
dtdx
dxdv
Velocity & Acceleration in Terms of x
If v = f(x);
2
2
2
21 v
dxd
dtxd
Proof:
dtdv
dtxd2
2
dtdx
dxdv
vdxdv
Velocity & Acceleration in Terms of x
If v = f(x);
2
2
2
21 v
dxd
dtxd
Proof:
dtdv
dtxd2
2
dtdx
dxdv
vdxdv
2
21 v
dvd
dxdv
Velocity & Acceleration in Terms of x
If v = f(x);
2
2
2
21 v
dxd
dtxd
Proof:
dtdv
dtxd2
2
dtdx
dxdv
vdxdv
2
21 v
dvd
dxdv
2
21 v
dxd
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 23
21 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 23
21 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 23
21 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 23
21 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 23
21 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 23
21 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 23
21 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 23
21 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 23
21 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 xx30 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 23
21 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 xx30 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 xv
dxd
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 xv
dxd
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 xv
dxd
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 xv
dxd
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 xv
dxd
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 xv
dxd
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 xv
dxd
0
1221 i.e.
2,1,0when
32
c
c
vxt
cxv 32
21
3
32
2
2
xv
xv
32xdtdx
23
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 xv
dxd
0
1221 i.e.
2,1,0when
32
c
c
vxt
cxv 32
21
3
32
2
2
xv
xv
32xdtdx
23
2x
(Choose –ve to satisfy the initial conditions)
23
21
xdxdt
23
21
xdxdt
cx
cx
cxt
22
22
1
21
21
when t = 0, x = 1
220 i.e.
cc
22 x
t
OR
x
dxxt1
23
21
x
x1
21
22
1
23
21
xdxdt
cx
cx
cxt
22
22
1
21
21
when t = 0, x = 1
220 i.e.
cc
22 x
t
OR
x
dxxt1
23
21
x
x1
21
22
1
112
x
23
21
xdxdt
cx
cx
cxt
22
22
1
21
21
when t = 0, x = 1
220 i.e.
cc
22 x
t
OR
x
dxxt1
23
21
x
x1
21
22
1
112
x
xt 22
23
21
xdxdt
cx
cx
cxt
22
22
1
21
21
when t = 0, x = 1
220 i.e.
cc
22 x
t
OR
x
dxxt1
23
21
x
x1
21
22
1
112
x
xt 22
222 t
x
23
21
xdxdt
cx
cx
cxt
22
22
1
21
21
when t = 0, x = 1
220 i.e.
cc
22 x
t
OR
x
dxxt1
23
21
x
x1
21
22
1
112
x
xt 22
222 t
x
222
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 22
21 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 22
21 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 22
21 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 22
21 32
cxxv 242
21
21
When x = 2, v = 5
1 125 16 42 2
12
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 22
21 32
cxxv 242
21
21
When x = 2, v = 5
1 125 16 42 2
12
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 22
21 32
cxxv 242
21
21
When x = 2, v = 5
1 125 16 42 2
12
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 22
21 32
cxxv 242
21
21
When x = 2, v = 5
1 125 16 42 2
12
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 22
21 32
cxxv 242
21
21
When x = 2, v = 5
1 125 16 42 2
12
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
12 xdtdx
xt
xdxdt
22
0 1
xxt 21tan
2tantan 11 xt
2tantan 11 tx
(ii) Hence find an expression for x in terms of t
12 xdtdx
xt
xdxdt
22
0 1
xxt 21tan
2tantan 11 xt
2tantan 11 tx
2tantan 1 tx
(ii) Hence find an expression for x in terms of t
12 xdtdx
xt
xdxdt
22
0 1
xxt 21tan
2tantan 11 xt
2tantan 11 tx
2tantan 1 tx
ttxtan21
2tan