Mathematics · Oct / Nov 2021 · Paper 51 · Question 3
YMA01
As printed in the paper
The question as text
3. A car is moving at a constant speed of 25 m s–1 along a straight horizontal road. The car is modelled as a particle. At time t = 0, the car is at the point A and the driver sees a road sign 48 m ahead. Let t seconds be the time that elapses after the car passes A. In a first model, the car is assumed to decelerate uniformly at 6 m s–2 from A until the car reaches the road sign.
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3(a)[2]
v =
2
25
2
−× ×
2 6 48
m s−)
1
v = 7 (
25 13
−
3(b)[4]
6
1
OR (25 13)
+
13
2
=
25
2
−×
2 6s
OR
25 2
× −
× ×
6 2
2
×
2
2
2
1
OR
13 2
× −
× −
( 6) 2
×
2
((s =) 38)
2
Total time = (48
−
38)
+ 2
13
36
10
13 =
213
(s) (2.76923…)
25 13
−
3(c)[4]
(2) (could be implied by 2.2)
6
1
(0.2 25)
(25 2
6 2 )
2
×
+
× −
× ×
(5 + 38)
2
[
]
48
−
(0.2 25)
×
+
38
Total time =
+ 0.2 + 2
13
168
38
=
2 65
(s) (2.58461538..)
65
Notes for question 3
3(a)
M1 Complete method to find v (condone sign errors)
A1 cao
3(b)
M1 Complete method to find time to reach 13
ms
1
M1 Complete method to find distance travelled in reaching 13
ms ft on
1
their 2 if necessary
DM1 Dependent on previous two M marks, Complete method to find the total
time, ft on their 2 and 38
A1 Correct answer. Allow 2.8 or better
M1 Complete method to find the time taken to reach 13
ms once it starts
1
3(c)
decelerating
ms ft
M1 Complete method to find total distance travelled in reaching 13
1
on their 2 if necessary
DM1 Dependent on previous two M marks, Complete method to find the total
time, ft on their 2 and 38
A1 Correct answer. Allow 2.6 or better
Allow column vectors throughout
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