Mathematics · May / June 2022 · Paper 31 · Question 8
YMA01
As printed in the paper
The question as text
8. v O T t Figure 4 Figure 4 is a graph showing the velocity of a sprinter during a 100 m race. –1, is modelled by the equation The sprinter’s velocity during the race, v m s v = 12 – e t – 10 – 12e –0.75t t 0 where t seconds is the time after the sprinter begins to run. According to the model,
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8(a)[5]
= −
e
t
−
10
+
9e
−
0.75
t
=
0
9
=
e
1.75
t−
10
or
9e
−
0.75
t
=
e
t
−
10
d
t
10
+
ln9
10
+
ln9
9
e
1.75
t
10
ln9
0.75
10
=
−
=
t
or
−
t
= −
t
=
t
1.75
1.75
t =
awrt 6.97
awrt 11.9 (ms1)
8(b)[4]
12 12e
−
−
0.75
t
−
e
(
t
−
10)
d
t
=
12
t
−
e
(
t
−
10)
+
16e
−
0.75
t
(
+
C
)
T
12
t
e
(
t
−
10)
16e
−
0.75
t
100
12
T
...
−
+
=
=
0
1 116 16e
(
)
T
=
−
−
0.75
T
+
e
T
−
10
−
e
−
10
*
12
8(c)[3]
(i)
1 116 16e
(
)
T
=
−
−
0.75 10
×
+
e
10 10
−
−
e
−
10
2
12
T =
awrt 9.7493
2
(ii)
9.7293 (seconds)
(a)
* Be aware this can be solved directly on a calculator. Calculus must be seen*
M1
Differentiates to
±
A
e
t
−
10
±
B
e
−
0.75
t
and sets equal to 0 Must be seen in (a) Do not accept
differentiating to eg
... et
t
−. Condone an expression of
10
12
±
A
e
t
−
10
±
B
e
−
0.75
t
only
A1
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