IBDP Maths AA: Topic SL 5.9: Kinematic problems: IB style Questions SL Paper 2

Question: [Maximum mark: 7]

A particle moves along a straight line so that its velocity, vms-1, after t seconds is given by v(t) = esint+ 4sint for 0 ≤ t ≤ 6.
(a) Find the value of t when the particle is at rest. 
(b) Find the acceleration of the particle when it changes direction. 
(c) Find the total distance travelled by the particle.

Answer/Explanation

Ans:

(a) recognizing at rest v = 0
t = 3.34692…
t = 3.35 (seconds)

Note: Award (M1)A0 for additional solutions to v = 0 eg t = −0.205 or t = 6.08.

(b) recognizing particle changes direction when v = 0 OR when t = 3.34692…
a = −4.71439…
a = −4.71(ms2)

Question: [Maximum mark: 7]

A particle moves in a straight line such that its velocity, vms-1, at time t seconds is given by \(v = \frac{\left ( t^{2}+1 \right )cos t}{4}, 0\leq t \leq 3.\)
(a) Determine when the particle changes its direction of motion.
(b) Find the times when the particle’s acceleration is -1.9ms-2.
(c) Find the particle’s acceleration when its speed is at its greatest.

Answer/Explanation

Ans:

(a) recognises the need to find the value of t when v = 0

(b) recognises that a(t) = v ‘(t)
t1 = 2.26277… , t2 = 2.95736…
t1 = 2.26 ,t2 = 2.96(s)

Note: Award M1A1A0 if the two correct answers are given with additional values outside 0 ≤ t ≤ 3 .

(c) speed is greatest at t = 3
a = −1.83778…
a = −1.84 (m s-2)

Question

A particle moves in a straight line. The velocity, v ms1 , of the particle at time t seconds is given by  v (t) = t sin t – 3, for 0 ≤ t ≤ 10.

The following diagram shows the graph of v .

                 

    1. Find the smallest value of t for which the particle is at rest. [2]

    2. Find the total distance travelled by the particle. [2]

    3. Find the acceleration of the particle when t = 7 . [2]

Answer/Explanation

Ans:

(a) recognising  v= o  t = 6.74416… = 6.74

= 37.0968…

= 37.1 (m)

(c)

recognizing acceleration at  t = 7  is given by v′(7)

acceleration = 5.93430… 

5.93

Question

The acceleration,  \(a{\text{ m}}{{\text{s}}^{ – 2}}\), of a particle at time t seconds is given by \[a = \frac{1}{t} + 3\sin 2t {\text{, for }} t \ge 1.\]

The particle is at rest when \(t = 1\) .

Find the velocity of the particle when \(t = 5\) .

Answer/Explanation

Markscheme

evidence of integrating the acceleration function     (M1)

e.g. \(\int {\left( {\frac{1}{t} + 3\sin 2t} \right)} {\rm{d}}t\)

correct expression \(\ln t – \frac{3}{2}\cos 2t + c\)     A1A1

evidence of substituting (1, 0)     (M1)

e.g. \(0 = \ln 1 – \frac{3}{2}\cos 2 + c\)

\(c = – 0.624\) \(\left( { = \frac{3}{2}\cos 2 – \ln {\text{1 or }}\frac{{\rm{3}}}{{\rm{2}}}\cos 2} \right)\)     (A1)

\(v = \ln t – \frac{3}{2}\cos 2t – 0.624\) \(\left( { = \ln t – \frac{3}{2}\cos 2t + \frac{3}{2}\cos {\text{2 or ln}}t – \frac{3}{2}\cos 2t + \frac{3}{2}\cos 2 – \ln 1} \right)\)     (A1)

\(v(5) = 2.24\) (accept the exact answer \(\ln 5 – 1.5\cos 10 + 1.5\cos 2\) ) A1     N3

[7 marks]

Question

A particle moves in a straight line with velocity \(v = 12t – 2{t^3} – 1\) , for \(t \ge 0\) , where v is in centimetres per second and t is in seconds.

Find the acceleration of the particle after 2.7 seconds.

[3]
a.

Find the displacement of the particle after 1.3 seconds.

[3]
b.
Answer/Explanation

Markscheme

recognizing that acceleration is the derivative of velocity (seen anywhere)     (R1)

e.g. \(a = \frac{{{{\rm{d}}^2}s}}{{{\rm{d}}{t^2}}},v’,12 – 6{t^2}\)

correctly substituting 2.7 into their expression for a (not into v)     (A1)

e.g. \(s”(2.7)\)

\({\text{acceleration}} = – 31.74\) (exact), \( – 31.7\)     A1     N3

[3 marks]

a.

recognizing that displacement is the integral of velocity     R1

e.g. \(s = \int v \)

correctly substituting 1.3     (A1)

e.g. \(\int_0^{1.3} {v{\rm{d}}t} \)

\({\text{displacement}} = 7.41195\) (exact), \(7.41\) (cm)     A1     N2

[3 marks]

b.

Question

The velocity \(v\) of a car, \(t\) seconds after leaving a fixed point \(A\), is given by
\(v=16-4e^{2t}\)

  1. Calculate the acceleration of the car after \(\ln 3\) seconds.
  2. Calculate the displacement of the car in terms of \(t\), given that the initial displacement is 0.
  3. The car is stationary at time \(t\). Find the exact value of \(t\).
  4. Find the displacement and the distance travelled after \(\ln 4\) seconds.
Answer/Explanation

Ans:

  1. \(a=-8e^{2t},\)
    At \(t= \ln 3 \space a=-8e^{2 \ln 3}=-72 ms^{-2}\)
  2. \(s=\int 16-4e^{2t}dt=16t-2e^{2t}+c\)
    \(s(0)=0\)⇒\(c=2\)
    \(s=16t-2e^{2t}+2\)
  3. \(v=0\)⇔\(16-4e^{2t}=0\)⇔\(t=\ln 2\)
  4. \(s=32 \ln 2 -30m,\)
    distance travelled = \(\int_{0}^{\ln 3}|16-4e^{2t}|dt\)
    =\(\int_{0}^{\ln 2}(16-4e^{2t})dt-\int_{\ln 2}^{\ln 3}(16-4e^{2t}dt\)
    =\(18m\)

NOTICE:
If the velocity is given in terms of time \(t\) then
\(a=\frac{dv}{dt}\)
If the velocity is given in terms of the displacement \(s\), then
\(a=\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=v\frac{dv}{ds}\)

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