Question 7:
--- 7(a) ---
7(a) | 3  3 
0.3
v= t2(+c) =0.2t2(+c) 
1.5   | M1 | For integration (do not penalise missing c)
The power of t must increase by 1 with a change of
coefficient. Use of v=at scores M0.
 8
Velocity = 1.6 = ms−1
 
 5 | A1 | ISW any extra work using the second equation for a.
2
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | −k − 1  − 1 
v= t 2+d  =2kt 2+d  
−0.5   | *M1 | For integration. No need for constant. Allow use of
given value of k=2.6.
The power of t must increase by 1 with a change of
coefficient. Use of v=at scores M0.
−k − 1  − 1 
Their1 .6= 4 2 +d =2k4 2 +d  Their1 .6=k+d 
−0.5  
−k − 1  − 1   k 
0.3= 16 2 +d=2k16 2 +d 0.3= +d
 
−0.5    2  | A1 FT | For both equations in k and d (Allow unsimplified) .
Attempt to solve for k or d | DM1 | Or substitute k = 2.6 into both equations and solve both
for d (with d 0).
Must get to ‘k=’ or ‘d=’.
− 1 −2.6 − 1
k = 2.6  v= 5.2t 2 −1 or  v= t 2 −1
−0.5 | A1 | AG (AG for k, not for the expression).
Allow unsimplified expression for v and/or in terms of
k.
If k is substituted then both equations must be shown
1
−
to give a value of d =−1 and getting v=5.2t 2 −1
SC B1 for solving the correct equations simultaneously
with no working seen and getting correct expression
for v.
SC A1 for correct expression for v if only the first M1
is scored.
Question | Answer | Marks | Guidance
7(b) | Alternative method for question 7(b) from using limits
−k − 1  − 1 
v= t 2+d  =2kt 2+d  
−0.5   | *M1 | For integration No need for constant. Allow use of
given value of k=2.6.
The power of t must increase by 1 with a change of
coefficient. Use of v=at scores M0.
−k − 1 −k − 1 − 1 − 1
4 2 − 16 2 =1.6−0.3 or 2k4 2 −2k16 2 =1.6−0.3
−0.5 −0.5 | A1 FT | OE For correct unsimplified equation in k from using
limits for v as their 1.6 and 0.3 and limits for t as 4 and
16.
Attempt to solve for k | DM1 | Must be an equation using 0.3 and their k, and 16 and
4, Must be subtracting the limits to form the equation
in k, but may have sign errors in their 1.6−0.3.
− 1 −2.6 − 1
k = 2.6, v=5.2t 2 −1 or v= t 2 −1
−0.5 | A1 | AG (AG for k, not for the expression).
Allow unsimplified expression for v and/or in terms of
k.
4
--- 7(c) ---
7(c) | 1
−
5.2T 2 −their1 =0 | M1 | For solving for T. Must get to ‘T=’. Must come from
integration, with their 1 from part (b) or found here not
equal to zero. Do not allow made up value of d.
676
T = or 27.04
25 | A1 | OE Must be exact. Allow both marks as long as
expression for v is correct, however obtained in
Q7(b).
2
Question | Answer | Marks | Guidance
--- 7(d) ---
7(d) | 4 3 5 27.04 1  1
0.2 − 5.2
0.2t2dt = t2 or  5.2t 2 −their1 dt = t2 −their1 t
 
2.5 0.5
 
0 4 | M1 | 3 1
−
For integration of 0.2t2 oe or 5.2t 2 −their 1 . May
be in terms of k. Not from any other expression. Their
1 may be zero (or replaced by zero).
Their 1 may came from either part (b) or part (c)
The power of t must increase by 1 with a change of
coefficient in at least one term.
4 27.04 4 27.04
 5  1   5  1 
0.2 5.2
= t2 + t2 −t =0.08t2 +10.4t2 −t 
 2.5   0.5       
0 4  0 4  | A1ft | For both integrals (unsimplified) No need for limits
FT non-zero value of d. May be in terms of k.
Their 1 may came from either part (b) or part (c).
=0.0832+(10.45.2−27.04)−(10.42−4) =2.56+10.24  | M1 | For correct use of limits (0 and 4 then 4 and their
27.04) in both of their integrals, which have come from
3 1
−
integration of 0.2t2 and 5.2t 2 −their1 . Not from any
other expression. Their 1 may be zero (or replaced by
zero).
Their 1 may came from either part (b) or part (c).
Allow M1 for d = 0 (the final answer is 35.8).
64
= or 12.8
5 | A1 | oe Awrt 12.8 Allow if using 27(.0) rather than 27.04
Allow all 4 marks as long as expression for v is
correct, however obtained in Q7(b).
Question | Answer | Marks | Guidance
7(d) | SC for using a calculator to integrate.
4 3
Either 0.2t2dt=2.56
0
27.04 1 
−
Or  5.2t 2 −1dt =10.24
 
 
4 | B1 | AWRT 2.56
Allow 10.2
Must use 27.04 or 27(.0) if latter integral.
Total distance = 12.8 m | B1 | AWRT 12.8. Allow if using 27(.0) rather than 27.04
Allow both B marks as long as expression for v is
correct, however obtained in Q7(b).
4