Math Repository
about
topic
al
year
ly
Yearly
2011
P1 Q8
Topical
DE
11 P1 Q8
2011 H2 Mathematics Paper 1 Question 8
Differential Equations (DEs)
Answers
(i)
1
20
ln
∣
10
+
v
10
−
v
∣
+
c
{\frac{1}{20} \ln \left| \frac{10 + v}{10 - v} \right| + c}
20
1
ln
10
−
v
10
+
v
+
c
(iia)
t
=
1
2
ln
(
10
+
v
10
−
v
)
{t = \frac{1}{2}\ln \left( \frac{10+v}{10-v} \right)}
t
=
2
1
ln
(
10
−
v
10
+
v
)
1
2
ln
3
s
{\frac{1}{2} \ln 3 \textrm{ s}}
2
1
ln
3
s
(iib)
7.62
m/s
{7.62 \textrm{ m/s}}
7.62
m/s
(iic)
The speed of the stone approaches
10
m/s
{10 \textrm{ m/s}}
10
m/s
for large values of
t
{t}
t
Full solutions
(i)
∫
1
100
−
v
2
d
v
=
1
2
(
10
)
ln
∣
10
+
v
10
−
v
∣
+
c
=
1
20
ln
∣
10
+
v
10
−
v
∣
+
c
■
\begin{align*} & \int \frac{1}{100 - v^2} \; \mathrm{d}v \\ &= \frac{1}{2(10)} \ln \left| \frac{10+v}{10-v} \right| + c \\ &= \frac{1}{20} \ln \left| \frac{10 + v}{10 - v} \right| + c \; \blacksquare \end{align*}
∫
100
−
v
2
1
d
v
=
2
(
10
)
1
ln
10
−
v
10
+
v
+
c
=
20
1
ln
10
−
v
10
+
v
+
c
■
(iia)
d
v
d
t
=
10
−
0.1
v
2
=
0.1
(
100
−
v
2
)
\begin{align*} \frac{\mathrm{d}v}{\mathrm{d}t} &= 10 - 0.1v^2 \\ &= 0.1 (100-v^2) \\ \end{align*}
d
t
d
v
=
10
−
0.1
v
2
=
0.1
(
100
−
v
2
)
∫
1
100
−
v
2
d
v
=
∫
0.1
d
t
1
20
ln
∣
10
+
v
10
−
v
∣
=
0.1
t
+
c
ln
∣
10
+
v
10
−
v
∣
=
2
t
+
20
c
∣
10
+
v
10
−
v
∣
=
e
2
t
+
20
c
10
+
v
10
−
v
=
±
e
20
c
e
2
t
=
A
e
2
t
\begin{align*} \int \frac{1}{100 - v^2} \; \mathrm{d}v &= \int 0.1 \; \mathrm{d}t \\ \frac{1}{20} \ln \left| \frac{10 + v}{10 - v} \right| &= 0.1t + c \\ \ln \left| \frac{10+v}{10-v} \right| &= 2t + 20c \\ \left| \frac{10+v}{10-v} \right| &= \mathrm{e}^{2t + 20c} \\ \frac{10+v}{10-v} &= \pm \mathrm{e}^{20c} \mathrm{e}^{2t} \\ &= A \mathrm{e}^{2t} \\ \end{align*}
∫
100
−
v
2
1
d
v
20
1
ln
10
−
v
10
+
v
ln
10
−
v
10
+
v
10
−
v
10
+
v
10
−
v
10
+
v
=
∫
0.1
d
t
=
0.1
t
+
c
=
2
t
+
20
c
=
e
2
t
+
20
c
=
±
e
20
c
e
2
t
=
A
e
2
t
When
t
=
0
,
{t=0, }
t
=
0
,
v
=
0
,
{v = 0,}
v
=
0
,
10
+
0
10
−
0
=
A
e
2
(
0
)
A
=
1
\begin{align*} \frac{10+0}{10-0} &= A \mathrm{e}^{2(0)} \\ A &= 1 \end{align*}
10
−
0
10
+
0
A
=
A
e
2
(
0
)
=
1
10
+
v
10
−
v
=
e
2
t
2
t
=
ln
(
10
+
v
10
−
v
)
t
=
1
2
ln
(
10
+
v
10
−
v
)
■
\begin{gather*} \frac{10+v}{10-v} = \mathrm{e}^{2t} \\ 2t = \ln \left( \frac{10+v}{10-v} \right) \\ t = \frac{1}{2}\ln \left( \frac{10+v}{10-v} \right) \; \blacksquare \end{gather*}
10
−
v
10
+
v
=
e
2
t
2
t
=
ln
(
10
−
v
10
+
v
)
t
=
2
1
ln
(
10
−
v
10
+
v
)
■
When
v
=
5
,
{v = 5,}
v
=
5
,
t
=
1
2
ln
(
10
+
5
10
−
5
)
=
1
2
ln
3
s
■
\begin{align*} t &= \frac{1}{2} \ln \left( \frac{10+5}{10-5} \right) \\ &= \frac{1}{2} \ln 3 \textrm{ s} \; \blacksquare \end{align*}
t
=
2
1
ln
(
10
−
5
10
+
5
)
=
2
1
ln
3
s
■
(iib)
10
+
v
10
−
v
=
e
2
t
10
+
v
=
(
10
−
v
)
e
2
t
v
(
1
+
e
2
t
)
=
10
e
2
t
−
10
\begin{gather*} \frac{10+v}{10-v} = \mathrm{e}^{2t} \\ 10+v = (10-v) \mathrm{e}^{2t} \\ v (1+\mathrm{e}^{2t}) = 10 \mathrm{e}^{2t} -10 \\ \end{gather*}
10
−
v
10
+
v
=
e
2
t
10
+
v
=
(
10
−
v
)
e
2
t
v
(
1
+
e
2
t
)
=
10
e
2
t
−
10
v
=
10
e
2
t
−
10
1
+
e
2
t
=
10
−
10
e
−
2
t
e
−
2
t
+
1
\begin{align*} v &= \frac{10 \mathrm{e}^{2t} -10}{1+\mathrm{e}^{2t}} \\ &= \frac{10-10\mathrm{e}^{-2t}}{\mathrm{e}^{-2t}+1} \end{align*}
v
=
1
+
e
2
t
10
e
2
t
−
10
=
e
−
2
t
+
1
10
−
10
e
−
2
t
When
t
=
1
,
{t=1,}
t
=
1
,
v
=
10
−
10
e
−
2
(
1
)
e
−
2
(
1
)
+
1
=
10
−
10
e
−
2
e
−
2
+
1
=
7.62
m/s (3 sf)
■
\begin{align*} v &= \frac{10-10\mathrm{e}^{-2(1)}}{\mathrm{e}^{-2(1)}+1} \\ &= \frac{10-10\mathrm{e}^{-2}}{\mathrm{e}^{-2}+1} \\ &= 7.62 \textrm{ m/s} \textrm{ (3 sf)} \; \blacksquare \end{align*}
v
=
e
−
2
(
1
)
+
1
10
−
10
e
−
2
(
1
)
=
e
−
2
+
1
10
−
10
e
−
2
=
7.62
m/s
(3 sf)
■
(iic)
As
t
→
∞
,
{t \to \infty, }
t
→
∞
,
e
−
2
t
→
0
{\mathrm{e}^{-2t} \to 0}
e
−
2
t
→
0
v
=
10
−
10
e
−
2
t
e
−
2
t
+
1
→
10
\begin{align*} v &= \frac{10-10\mathrm{e}^{-2t}}{\mathrm{e}^{-2t}+1} \\ &\to 10 \end{align*}
v
=
e
−
2
t
+
1
10
−
10
e
−
2
t
→
10
Hence the speed of the stone approaches
10
m/s
{10 \textrm{ m/s}}
10
m/s
for large values of
t
■
{t \; \blacksquare}
t
■
Back to top ▲