Question 20:
--- 20(a) ---
20(a) | Models the rate of change of
depth using an equation of the
d h
form =  k ( h − 5 )
d t | 3.3 | M1 | dh
=−k(h−5 )
dt
dh
when h=130, =−1.5
dt
−1.5=−k125
k =0.012
dh
=−0.012 (h−5 )
dt
Substitutes h = 130
d h
and =  1 .5 into
d t
d h
=  k ( h − 5 )
d t | 3.1b | M1
Completes argument with no
sign slips to show
dh
=−0.012 (h−5 )
dt | 2.1 | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 20(b) ---
20(b) | Separates variables to obtain an
equation of the form
A
 d h =  B d t
h − 5
PI by l n ( h − 5 ) = − 0 .0 1 2 t OE | 3.1a | M1 | 1 dh
=−0.012
h−5 dt
1
 dh=−0.012 dt
h−5
(h−5 )=−0.012t+c
ln
h−5= Ae −0.012t
t =0,h=130 A=125
h=5+125e −0.012t
Integrates one of their integrals
of the form
A
 dh or  B d t correctly.
h−5
ln(h−5 )=−0.012tOE
PI by | 1.1a | M1
Obtains correct integrated
equation.
Condone missing + c | 1.1b | A1
Uses t = 0 , h = 1 3 0 to obtain
their constant of integration. | 3.1b | M1
Obtains 5 + 1 2 5 e − 0 .0 1 2 t
OE
Accept 5 + e − 0 .0 1 2 t + p where
p =ln125 or AWRT 4.83 | 3.3 | A1
Subtotal | 5
Q | Marking instructions | AO | Marks | Typical solution
--- 20(c) ---
20(c) | Uses h = 65 in their answer from
part (b)
and obtains a final positive value | 3.4 | M1 | 5 + 1 2 5 − e 0 .0 1 2 t = 6 5
t = 6 1 .1 6 4
61 minutes
Obtains AWRT 61 minutes
Accept 62 minutes
Must have correct units | 3.2a | A1
Subtotal | 2
Question 20 Total | 10
Question Paper Total | 100