Edexcel A level Maths (9MA0) · Mechanics › Kinematics
Practise Kinematics graphs. 1 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 cyclist starts from rest and accelerates uniformly to a speed of 12 m s−1 in 4 s. The cyclist then travels at 12 m s−1 for 10 s, and then decelerates uniformly to rest in 6 s. The velocity–time graph shows the motion.
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(a) A student says that the distance travelled in the first 4 seconds is \(12 \div 4 = 3\) m. Explain the student's error and find the correct distance.[2]
(b) Find the total distance travelled by the cyclist.[2]
(c) Find the deceleration of the cyclist in the final 6 seconds.[1]
Show the answer and mark scheme
(a)Answer: 24 m
B1 for \(12 \div 4\) is the gradient, which gives the acceleration (3 m s−2), not the distance; the distance is the area under the graph
B1 for 24 (m)
Worked solution: \(12 \div 4 = 3\) is the gradient of the graph, which is the acceleration (3 m s−2). Distance is the area under the graph: \(\frac{1}{2} \times 4 \times 12 = 24\) m.
(b)Answer: 180 m
M1 for the area of the trapezium, e.g. \(\frac{1}{2}(10 + 20) \times 12\)
A1 for 180 (m)
Worked solution: Area \(= \frac{1}{2}(10 + 20) \times 12 = 180\) m (or \(24 + 120 + 36\)).
(c)Answer: 2 m s−2
B1 for 2 (m s−2)
Worked solution: Deceleration \(= \frac{12}{6} = 2\) m s−2.
Question 2Medium8 marks
A cyclist moves along a straight horizontal path. Starting from rest at the point \(A\), the cyclist accelerates uniformly for \(T\) seconds, reaching a speed of 6 m s−1. The cyclist then travels at this speed for \(2T\) seconds, and then decelerates uniformly for \(2T\) seconds, coming to rest at the point \(B\). The distance \(AB\) is 126 m.
(a) Sketch a speed–time graph for the motion of the cyclist from \(A\) to \(B\).[2]
(b) Find the value of \(T\).[4]
(c) Find the acceleration of the cyclist during the first stage of the motion.[2]
Show the answer and mark scheme
(a)Answer: A trapezium: a straight line from \((0, 0)\) up to \((T, 6)\), horizontal to \((3T, 6)\), then a straight line down to \((5T, 0)\).
B1 for a trapezium shape starting at the origin and finishing on the \(t\)-axis
B1 for 6, \(T\), \(3T\) and \(5T\) marked correctly
Worked solution: The graph rises in a straight line from the origin to speed 6 at \(t = T\), is horizontal until \(t = 3T\), and then falls in a straight line to zero at \(t = 5T\).
(b)Answer: \(T = 6\)
M1 for using the area under their graph to form an equation in \(T\)
A1 for a correct equation, e.g. \(\tfrac{1}{2}(2T + 5T) \times 6 = 126\)
dM1 for solving their linear equation
A1 for \(T = 6\)
Worked solution: Area of the trapezium = distance: \(\tfrac{1}{2}(2T + 5T) \times 6 = 126\) \(21T = 126 \Rightarrow T = 6\)
(c)Answer: \(1\) m s−2
M1 for \(\frac{6}{\text{their }T}\)
A1ft for 1 (m s−2)
Worked solution: \(a = \frac{6}{6} = 1\text{ m s}^{-2}\)
Question 3Hard12 marks
Two cars, \(P\) and \(Q\), move along a straight horizontal road in the same direction. At time \(t = 0\), car \(P\) starts from rest at the point \(O\) and accelerates uniformly at 1.25 m s−2 until it reaches a speed of 10 m s−1. It then moves at this constant speed. Also at \(t = 0\), car \(Q\) passes \(O\) moving at a constant speed of 8 m s−1.
(a) Sketch, on the same axes, the velocity–time graphs for the two cars, for the interval from \(t = 0\) until car \(P\) overtakes car \(Q\).[3]
(b) Find the time at which car \(P\) overtakes car \(Q\).[5]
(c) Find the distance from \(O\) at which car \(P\) overtakes car \(Q\).[1]
(d) Find the greatest distance between the two cars before car \(P\) overtakes car \(Q\).[3]
Show the answer and mark scheme
(a)Answer: \(P\): a straight line from the origin to \((8, 10)\), then horizontal at 10. \(Q\): a horizontal line at 8, starting on the velocity axis. The lines cross at \(t = 6.4\).
B1 for the graph of \(P\): a line through the origin then horizontal
B1 for the graph of \(Q\): a horizontal line starting on the velocity axis, below the horizontal part of \(P\)'s graph
B1 for 10 and 8 marked on the velocity axis and 8 on the time axis
Worked solution: Car \(P\) reaches 10 m s−1 at \(t = \frac{10}{1.25} = 8\). Its graph is a straight line from the origin to \((8, 10)\) and then horizontal. Car \(Q\)'s graph is a horizontal line at 8.
(b)Answer: \(t = 20\) s
M1 for the distance travelled by \(Q\) as \(8t\)
M1 for the distance travelled by \(P\) as the area under its graph, e.g. \(\tfrac{1}{2} \times 8 \times 10 + 10(t - 8)\)
M1 for equating the two distances
A1 for a correct equation, e.g. \(10t - 40 = 8t\)
A1 for \(t = 20\)
Worked solution: \(P\) reaches its top speed at \(t = 8\), having travelled \(\tfrac{1}{2} \times 8 \times 10 = 40\) m, while \(Q\) has travelled \(64\) m; so \(P\) has not yet caught up. For \(t \gt 8\): \(40 + 10(t - 8) = 8t \Rightarrow 2t = 40 \Rightarrow t = 20\)
(c)Answer: 160 m
B1ft for 160 (m)
Worked solution: \(8 \times 20 = 160\) m
(d)Answer: \(25.6\) m
M1 for recognising that the gap is greatest when the two speeds are equal, at \(t = 6.4\)
M1 for the difference in the areas under the two graphs up to this time
A1 for 25.6 (m)
Worked solution: The gap grows while \(Q\) is faster and shrinks once \(P\) is faster, so it is greatest when both speeds equal 8 m s−1, at \(t = \frac{8}{1.25} = 6.4\). Gap = \(8 \times 6.4 - \tfrac{1}{2} \times 6.4 \times 8 = 25.6\) m (the area of the triangle between the two graphs).