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Date May 2021 Marks available 2 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, ams-2, of a particle moving in a horizontal line at time t seconds, t0, is given by a=-(1+v) where vms-1 is the particle’s velocity and v>-1.

At t=0, the particle is at a fixed origin O and has initial velocity v0ms-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.

[6]
a.

Show that the time T taken for the particle to reach smax satisfies the equation eT=1+v0.

[2]
b.i.

By solving an appropriate differential equation and using the result from part (b) (i), find an expression for smax in terms of v0.

[5]
b.ii.

By using the result to part (b) (i), show that v(T-k)=ek-1.

[2]
c.

Deduce a similar expression for v(T+k) in terms of k.

[2]
d.

Hence, show that v(T-k)+v(T+k)0.

[3]
e.

Markscheme

dvdt=-(1+v)         (A1)

1dt=-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, v=v0        M1

C=ln(1+v0)

t=ln(1+v0)-ln(1+v)

t=ln(1+v01+v)et=1+v01+v        A1

et(1+v)=1+v0

1+v=(1+v0)e-t        A1

v(t)=(1+v0)e-t-1        AG

 

OR

t-C=-ln(1+v)et-C=1(1+v)

Attempt to find C with initial conditions t=0, v=v0        M1

e-C=1(1+v0)C=ln(1+v0)

t-ln(1+v0)=-ln(1+v)t=ln(1+v0)-ln(1+v)

t=ln(1+v01+v)et=1+v01+v        A1

et(1+v)=1+v0

1+v=(1+v0)e-t        A1

v(t)=(1+v0)e-t-1        AG

 

OR

t-C=-ln(1+v)e-t+C=1+v        A1

ke-t-1=v

Attempt to find k with initial conditions t=0, v=v0        M1

k=1+v0

e-t(1+v0)=1+v        A1

v(t)=(1+v0)e-t-1        AG

 

Note: condone use of modulus within the ln function(s)

 

[6 marks]

a.

recognition that when t=T, v=0        M1

(1+v0)e-T-1=0e-T=11+v0        A1

eT=1+v0        AG

 

Note: Award M1A0 for substituting v0=eT-1 into v and showing that v=0.

 

[6 marks]

b.i.

s(t)=v(t)dt (=((1+v0)e-t-1)dt)        (M1)

=-(1+v0)e-t-t(+D)        A1

(t=0, s=0 so) D=1+v0        A1

s(t)=-(1+v0)e-t-t+1+v0

at smax, eT=1+v0T=ln(1+v0)

Substituting into s(t)(=-(1+v0)e-t-t+1+v0)        M1

smax=-(1+v0)(11+v0)-ln(1+v0)+v0+1        A1

(smax=v0-ln(1+v0))

 

[5 marks]

b.ii.

METHOD 1

v(T-k)=(1+v0)e-Tek-1        (M1)

=(1+v0)(11+v0)ek-1        A1

=ek-1        AG

 

METHOD 2

v(T-k)=(1+v0)e-(T-k)-1

=eTe-(T-k)-1         M1

=eT-T+k-1        A1

=ek-1        AG

 

[2 marks]

c.

METHOD 1

v(T+k)=(1+v0)e-Te-k-1        (A1)

=e-k-1       A1

 

METHOD 2

v(T+k)=(1+v0)e-(T+k)-1        (A1)

        =eTe-(T+k)-1

=eT-T-k-1

=e-k-1       A1

 

[2 marks]

d.

METHOD 1

v(T-k)+v(T+k)=ek+e-k-2       A1

attempt to express as a square       M1

=(ek2-e-k2)2 (0)       A1

so v(T-k)+v(T+k)0       AG

 

METHOD 2

v(T-k)+v(T+k)=ek+e-k-2       A1

Attempt to solve ddk(ek+e-k)=0  (k=0)       M1

minimum value of 2, (when k=0), hence ek+e-k2       R1

so v(T-k)+v(T+k)0       AG

 

[3 marks]

e.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.
[N/A]
d.
[N/A]
e.

Syllabus sections

Topic 5 —Calculus » SL 5.9—Kinematics problems
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Topic 5 —Calculus » AHL 5.18—1st order DE’s – Euler method, variables separable, integrating factor, homogeneous DE using sub y=vx
Topic 5 —Calculus

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