Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | Attempt to integrate 12−2t | M1 | For integration, the power of t must increase
by 1 in at least 1 term with a change of
coefficient in the same term.
No +c required for this mark.
s=vt is M0.
 2t2 
v =12t− (+c) =12t−t2(+c)
 
 2  | A1 | No +c required for this mark.
Allow unsimplified.
Use boundary conditions to get c=−20 | B1
Solve 12t−t2 −20=0 to get t =2 and t =10 | B1 | soi
Correct graph inverted quadratic starting at ( 0, −20 ) and ending at ( 12, −20 ) | B1 | Ignore anything outside 0 t 12.
t =2 and t =10 need not be shown.
5
Question | Answer | Marks | Guidance
--- 5(b) ---
5(b) | Attempt to integrate their 12t−t2 −20 | *M1 | Integrating their 2 or 3 term expression for v
from (a) which has come from integration.
For integration, the power of t must increase
by 1 in at least 1 term with a change of
coefficient in the same term.
12 1  1
s= t2 − t3−20t(+d)= 6t2 − t3−20t (+d)
 
 2 3  3 | A1ft | ft their +c0. Allow unsimplified.
 1 
Attempt to evaluate their 6t2 − t3−20t for any of t =0 to t =their 2 or
 
 3 
t =their 2 to t =their1 0 or t =their1 0 to t =12 | DM1 | Correct use of correct limits for one time
interval.
 1 
Attempt to evaluate their 6t2 − t3−20t for all of t =0 to t =their 2 or
 
 3 
t =their 2 to t =their1 0 or t =their1 0 to t =12 | DM1 | Correct use of correct limits for all 3 time
intervals, ignore signs here.
 56  200  56  200 368
s=−  − −0   +  −  −  − 48−  = 123 m
 3   3  3   3  3 | A1 | Awrt 123
Question | Answer | Marks | Guidance
5(b) | 2 1  56
Either s=  6t2 − t3−20t  dt = =18.7
 3  3
0
10 1  256
Or s=  6t2 − t3−20t  dt = =85.3
 3  3
2
12 1  56
Or s=  6t2 − t3−20t  dt = =18.7
 3  3
10 | B1 | 20
Allow  0.25t2 −8t+60 dt =26.
10
56 256 56 368
s= + + = 123 m
 
 3 3 3  3 | B1 | Awrt 123
12 1 368
Allow s=  6t2 − t3−20t dt = 123
3 3
0
m for B2.
5
Question | Answer | Marks | Guidance