Edexcel A level Maths (9MA0) · Mechanics › Kinematics
Practise Variable acceleration using calculus. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
A particle \(P\) moves along a straight line. At time \(t\) seconds, \(t \ge 0\), the displacement of \(P\) from a fixed point \(O\) on the line is \(s\) metres, where\[s = t^{3} + 5t^{2} + 2t\]
(a) Find an expression for the velocity of \(P\) at time \(t\) seconds.[2]
(b) Find the velocity and the acceleration of \(P\) when \(t = 4\).[3]
Show the answer and mark scheme
(a)Answer: \(v = 3t^{2} + 10t + 2\)
M1 for differentiating: at least one power of \(t\) reduced by 1
(b)Answer: Velocity \(90\) m s−1; acceleration \(34\) m s−2
B1ft for velocity \(90\) (m s−1)
M1 for differentiating their \(v\) to obtain \(a = 6t + 10\)
A1 for acceleration \(34\) (m s−2)
Worked solution: \(v(4) = 90\text{ m s}^{-1}\) \(a = \frac{\mathrm{d}v}{\mathrm{d}t} = 6t + 10 \Rightarrow a(4) = 34\text{ m s}^{-2}\)
Question 2Medium8 marks
A particle \(P\) moves along a straight line. At time \(t\) seconds, \(t \ge 0\), the velocity of \(P\) is \(v\) m s−1, where \[v = 3t^2 - 12t + 9\]A student is asked to find the distance travelled by \(P\) in the first 4 seconds and writes: \(\int_{0}^{4} (3t^2 - 12t + 9)\,\mathrm{d}t = \left[t^3 - 6t^2 + 9t\right]_{0}^{4} = 64 - 96 + 36 = 4\) m
(a) Explain why the student's answer is not the distance travelled.[2]
(b) Find the total distance travelled by \(P\) in the first 4 seconds.[4]
(c) Find the greatest speed of \(P\) for \(1 \le t \le 3\).[2]
Show the answer and mark scheme
(a)Answer: \(v\) changes sign at \(t = 1\) and \(t = 3\), so \(P\) changes direction; the integral gives the displacement, in which motion in the negative direction cancels motion in the positive direction.
B1 for \(v = 0\) at \(t = 1\) and \(t = 3\) (\(v \lt 0\) between them), so \(P\) changes direction
B1 for the integral gives the (net) displacement, not the distance
Worked solution: \(v = 3(t - 1)(t - 3)\), which is negative for \(1 \lt t \lt 3\): \(P\) moves backwards during that time. The integral gives the displacement, in which the backward motion is subtracted from the forward motion.
(b)Answer: 12 m
M1 for solving \(v = 0\): \(t = 1\) and \(t = 3\)
M1 for evaluating the displacement function \(s = t^3 - 6t^2 + 9t\) at \(t = 1\), 3 and 4 (4, 0, 4)
M1 for adding the distances for each interval: \(4 + |0 - 4| + 4\)
M1 for \(\frac{\mathrm{d}v}{\mathrm{d}t} = 6t - 12 = 0 \Rightarrow t = 2\)
A1 for speed 3 (m s−1) (from \(v = -3\))
Worked solution: \(\frac{\mathrm{d}v}{\mathrm{d}t} = 6t - 12 = 0\) at \(t = 2\), where \(v = 12 - 24 + 9 = -3\). At \(t = 1\) and \(t = 3\), \(v = 0\). So the greatest speed is 3 m s−1.
Question 3Hard11 marks
A particle \(P\) moves along a straight line. At time \(t\) seconds, \(t \ge 0\), the displacement of \(P\) from a fixed point \(O\) on the line is \(s\) metres, where\[s = 2t^{3} - 15t^{2} + 24t\] In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
(a) Find the values of \(t\) for which \(P\) is instantaneously at rest.[3]
(b) Find the total distance travelled by \(P\) in the interval \(0 \le t \le 6\).[5]
(c) Find the minimum value of \(v\), justifying your answer.[3]
Show the answer and mark scheme
(a)Answer: \(t = 1\) and \(t = 4\)
M1 for differentiating: \(v = 6t^{2} - 30t + 24\)
M1 for setting their \(v = 0\) and solving a 3-term quadratic
A1 for \(t = 1\) and \(t = 4\)
Worked solution: \(v = 6t^{2} - 30t + 24 = 6(t - 1)(t - 4) = 0 \Rightarrow t = 1 \text{ or } t = 4\)
(b)Answer: \(90\) m
M1 for recognising that \(P\) changes direction at \(t = 1\) and \(t = 4\) (splitting the interval)
M1 for finding \(s\) at \(t = 1\), \(t = 4\) and \(t = 6\)
A1 for \(s(1) = 11\), \(s(4) = -16\) and \(s(6) = 36\)
M1 for adding the magnitudes of the three displacements
B1 for a justification that this is a minimum, e.g. \(\frac{\mathrm{d}^2v}{\mathrm{d}t^2} = 12 \gt 0\) or \(v\) is a positive quadratic in \(t\)
Worked solution: \(\frac{\mathrm{d}v}{\mathrm{d}t} = 12t - 30 = 0 \Rightarrow t = 2.5\), and \(v(2.5) = -13.5\). \(\frac{\mathrm{d}^2v}{\mathrm{d}t^2} = 12 \gt 0\), so this is a minimum: the least value of \(v\) is \(-13.5\) m s−1.