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 ms–1 , 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 .
Find the smallest value of t for which the particle is at rest. [2]
Find the total distance travelled by the particle. [2]
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.
Find the displacement of the particle after 1.3 seconds.
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]
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]
Question
The velocity \(v\) of a car, \(t\) seconds after leaving a fixed point \(A\), is given by
\(v=16-4e^{2t}\)
- Calculate the acceleration of the car after \(\ln 3\) seconds.
- Calculate the displacement of the car in terms of \(t\), given that the initial displacement is 0.
- The car is stationary at time \(t\). Find the exact value of \(t\).
- Find the displacement and the distance travelled after \(\ln 4\) seconds.
Answer/Explanation
Ans:
- \(a=-8e^{2t},\)
At \(t= \ln 3 \space a=-8e^{2 \ln 3}=-72 ms^{-2}\) - \(s=\int 16-4e^{2t}dt=16t-2e^{2t}+c\)
\(s(0)=0\)⇒\(c=2\)
\(s=16t-2e^{2t}+2\) - \(v=0\)⇔\(16-4e^{2t}=0\)⇔\(t=\ln 2\)
- \(s=32 \ln 2 -30m,\)
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
\(a=\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=v\frac{dv}{ds}\)