Date | May 2021 | Marks available | 6 | Reference code | 21M.1.AHL.TZ2.11 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 2 |
Command term | Show that | Question number | 11 | Adapted from | N/A |
Question
The acceleration, a ms-2, of a particle moving in a horizontal line at time t seconds, t≥0, is given by a=-(1+v) where v ms-1 is the particle’s velocity and v>-1.
At t=0, the particle is at a fixed origin O and has initial velocity v0 ms-1.
Initially at O, the particle moves in the positive direction until it reaches its maximum displacement from O. The particle then returns to O.
Let s metres represent the particle’s displacement from O and smax its maximum displacement from O.
Let v(T-k) represent the particle’s velocity k seconds before it reaches smax, where
v(T-k)=(1+v0)e-(T-k)-1.
Similarly, let v(T+k) represent the particle’s velocity k seconds after it reaches smax.
By solving an appropriate differential equation, show that the particle’s velocity at time t is given by v(t)=(1+v0)e-t-1.
Show that the time T taken for the particle to reach smax satisfies the equation eT=1+v0.
By solving an appropriate differential equation and using the result from part (b) (i), find an expression for smax in terms of v0.
By using the result to part (b) (i), show that v(T-k)=ek-1.
Deduce a similar expression for v(T+k) in terms of k.
Hence, show that v(T-k)+v(T+k)≥0.
Markscheme
dvdt=-(1+v) (A1)
∫1 dt=∫-11+vdv (or equivalent / use of integrating factor) M1
t=-ln(1+v)(+C) A1
EITHER
attempt to find C with initial conditions t=0, M1
A1
A1
AG
OR
Attempt to find with initial conditions M1
A1
A1
AG
OR
A1
Attempt to find with initial conditions M1
A1
AG
Note: condone use of modulus within the ln function(s)
[6 marks]
recognition that when M1
A1
AG
Note: Award M1A0 for substituting into and showing that .
[6 marks]
(M1)
A1
( so) A1
at
Substituting into M1
A1
[5 marks]
METHOD 1
(M1)
A1
AG
METHOD 2
M1
A1
AG
[2 marks]
METHOD 1
(A1)
A1
METHOD 2
(A1)
A1
[2 marks]
METHOD 1
A1
attempt to express as a square M1
A1
so AG
METHOD 2
A1
Attempt to solve M1
minimum value of , (when ), hence R1
so AG
[3 marks]