(a)State what the gradient of a velocity-time graph represents.(1)
(b)State what the area between a velocity-time graph and the time axis represents.(1)
(Total for Question 1 is 2 marks)
2
The velocity-time graph shows the motion of a cyclist over 14 seconds. The cyclist accelerates uniformly from rest to a velocity of 8 m/s in the first 4 seconds, travels at a constant velocity of 8 m/s until t = 10 seconds, and then decelerates uniformly to rest by t = 14 seconds.
(a)Write down the velocity of the cyclist at t = 6 seconds.(1)
(b)Write down the time interval during which the cyclist travels at a constant velocity.(1)
(c)Describe what is happening to the speed of the cyclist between t = 10 and t = 14 seconds.(1)
(Total for Question 2 is 3 marks)
3
A car accelerates uniformly from rest to a velocity of 15 m/s in 6 seconds. Later in its journey, the car decelerates uniformly from 15 m/s to rest in 10 seconds.
(a)Calculate the acceleration of the car during the first 6 seconds.(2)
(b)Calculate the deceleration of the car while it slows to rest.(2)
(Total for Question 3 is 4 marks)
4
A motorbike's velocity-time graph is a straight line from (0, 0) to (10, 25), where time is in seconds and velocity is in m/s.
(Total for Question 4 is 2 marks)
5
The velocity-time graph shows the motion of a sprinter over 10 seconds. The sprinter accelerates uniformly from rest to a velocity of 9 m/s in the first 3 seconds, then runs at a constant velocity of 9 m/s for the remaining 7 seconds.
(a)Calculate the distance travelled by the sprinter during the first 3 seconds.(2)
(b)Calculate the total distance travelled by the sprinter in the 10 seconds shown on the graph.(3)
(Total for Question 5 is 5 marks)
6
A train travelling at 24 m/s decelerates uniformly at a rate of 1.6 m/s2 until it comes to rest.
(a)Calculate the time taken for the train to come to rest.(2)
(b)Calculate the distance travelled by the train while it is decelerating.(3)
(Total for Question 6 is 5 marks)
7
A car's velocity increases uniformly from 36 km/h to 90 km/h in 5 seconds.
(a)Show that the initial velocity of the car is 10 m/s.(1)
(b)Calculate the acceleration of the car, giving your answer in m/s2.(3)
(Total for Question 7 is 4 marks)
8
The velocity-time graph shows the motion of a delivery van over 20 seconds. The van accelerates uniformly from rest to a velocity of 20 m/s in the first 6 seconds, travels at this constant velocity for the next 9 seconds, and then decelerates uniformly to rest over the final 5 seconds.
(a)Calculate the total distance travelled by the van in the 20 seconds shown.(4)
(b)Calculate the average speed of the van for the whole 20 second journey.(2)
(Total for Question 8 is 6 marks)
9
A cyclist accelerates uniformly from rest to reach a velocity of V m/s in 8 seconds, then travels at this constant velocity for a further 12 seconds. The total distance travelled by the cyclist in these 20 seconds is 224 metres.
(a)Form an equation in V and solve it to find the value of V.(4)
(b)Hence calculate the acceleration of the cyclist during the first 8 seconds.(2)
(Total for Question 9 is 6 marks)
10
A bus leaves a stop and accelerates uniformly to a speed of 12 m/s in 15 seconds. It then travels at 12 m/s for 40 seconds, before decelerating uniformly to rest in 10 seconds. The bus then waits at a second stop for 20 seconds, before accelerating uniformly back up to 12 m/s over the next 12 seconds.
(Total for Question 10 is 4 marks)
11
Priya and Tom start cycling from the same point at the same time and travel along the same straight road for 20 seconds. Priya accelerates uniformly from rest to a velocity of 10 m/s in the first 5 seconds, then travels at this constant velocity for the remaining 15 seconds. Tom cycles at a constant velocity of 8 m/s for the whole 20 seconds.
(a)Calculate the distance Priya has travelled after 20 seconds.(3)
(b)Calculate the distance Tom has travelled after 20 seconds.(1)
(c)Determine which cyclist is further along the road after 20 seconds, and by how much.(2)
(Total for Question 11 is 6 marks)
12
A stone is thrown vertically upwards. It leaves the thrower's hand with a velocity of 15 m/s and returns to the same height 3 seconds later. Taking upward as the positive direction, and ignoring air resistance, consider the velocity-time graph of the stone's motion for these 3 seconds.
(a)Which velocity-time graph correctly represents the motion of the stone?(1)
A) A horizontal line at velocity = 15 m/s for the whole 3 seconds
B) A straight line from (0, 15) to (3, -15), passing through (1.5, 0)
C) A straight line from (0, 15) to (3, 0)
D) A curve that decreases from (0, 15) to a minimum and then increases back to (3, 15)
(b)State the time at which the stone is at its highest point, and explain how this can be seen from the graph.(2)
(Total for Question 12 is 3 marks)
13
A remote-control car moves along a straight track. Taking the direction it initially moves in as positive, its velocity-time graph is as follows: from t = 0 to t = 4 seconds, velocity increases uniformly from 0 to 6 m/s; from t = 4 to t = 6 seconds, velocity is constant at 6 m/s; from t = 6 to t = 10 seconds, velocity decreases uniformly from 6 m/s to -4 m/s; from t = 10 to t = 12 seconds, velocity is constant at -4 m/s.
(a)State the time at which the car changes direction.(1)
(b)Calculate the total distance travelled by the car in the first 12 seconds.(5)
(c)Calculate the displacement of the car from its starting point after 12 seconds.(2)
(Total for Question 13 is 8 marks)
14
A ball is dropped and falls vertically. Its downward velocity is recorded every second for 5 seconds, as air resistance means the ball's acceleration decreases over time. The results are shown in the table: t (s): 0 1 2 3 4 5 v (m/s): 0 8.5 15.0 19.0 21.0 22.0
(a)Use the trapezium rule, with all 5 strips of width 1 second, to estimate the total distance fallen by the ball in the 5 seconds.(4)
(b)State whether the trapezium rule gives an overestimate or an underestimate of the true distance fallen, giving a reason for your answer.(2)
(Total for Question 14 is 6 marks)
15
For the ball in Question 14, a tangent is drawn to the velocity-time curve at t = 2 seconds. The tangent passes through the points (0, 5) and (4, 25), which lie on the tangent line (these are not data points from the table).
(a)Use the tangent to estimate the acceleration of the ball at t = 2 seconds.(3)
(b)Explain why this method gives only an estimate, rather than an exact value, of the acceleration at t = 2 seconds.(1)
(Total for Question 15 is 4 marks)
16
A particle's velocity-time graph has three phases. In the first phase, the particle accelerates uniformly at 4 m/s2 from an initial velocity of u m/s to 30 m/s, taking 5 seconds. In the second phase, it travels at a constant 30 m/s for 20 seconds. In the third phase, it decelerates uniformly from 30 m/s to rest in t seconds. The total distance travelled by the particle over all three phases is 850 metres.
(a)Find the value of u.(3)
(b)Find the value of t.(4)
(Total for Question 16 is 7 marks)
Mark scheme · 8.7 Velocity Time Graphs
Question 1
(a) B1 acceleration oe (rate of change of velocity)
(a) Answer: Acceleration
(b) B1 distance travelled oe
(b) Answer: Distance travelled
Question 2
(a) B1 8 (m/s)
(a) Answer: 8 m/s
(b) B1 4 ≤ t ≤ 10 oe (between t = 4 and t = 10 seconds)
(b) Answer: From t = 4 seconds to t = 10 seconds
(c) B1 decelerating oe (slowing down, to rest by t = 14)
(c) Answer: The cyclist is decelerating (slowing down) to rest.
Question 3
(a) M1 15 / 6
(a) A1 2.5 (m/s2) cao
(a) Answer: 2.5 m/s2
(b) M1 15 / 10
(b) A1 1.5 (m/s2) cao
(b) Answer: 1.5 m/s2
Question 4
M1 gradient = (25 - 0) / (10 - 0)
A1 cso: 2.5 (m/s2), with working shown (answer is given)
Answer: 2.5 m/s2
Question 5
(a) M1 0.5 x 3 x 9
(a) A1 13.5 (m) cao
(a) Answer: 13.5 m
(b) M1 9 x 7 (area of the constant-velocity rectangle)
(b) M1 13.5 + 63 (adds triangle and rectangle areas, ft from (a))
(b) A1 76.5 (m) cao
(b) Answer: 76.5 m
Question 6
(a) M1 24 / 1.6
(a) A1 15 (s) cao
(a) Answer: 15 s
(b) M1 correct method for area under graph, e.g. 0.5 x (24 + 0) x 15 or 0.5 x 24 x 15 (ft time from (a))
(b) M1 correct substitution of values
(b) A1 180 (m) cao
(b) Answer: 180 m
Question 7
(a) B1 cso: 36 x 1000 / 3600 = 10 (m/s), with working shown
(a) Answer: 10 m/s
(b) M1 converts 90 km/h to 25 m/s
(b) M1 (25 - 10) / 5
(b) A1 3 (m/s2) cao
(b) Answer: 3 m/s2
Question 8
(a) M1 0.5 x 6 x 20 (area of first triangle, = 60)
(a) M1 20 x 9 (area of rectangle, = 180)
(a) M1 0.5 x 5 x 20 (area of final triangle, = 50), and adds all three areas
(a) A1 290 (m) cao
(a) Answer: 290 m
(b) M1 290 / 20 (ft from (a))
(b) A1 14.5 (m/s) cao
(b) Answer: 14.5 m/s
Question 9
(a) M1 0.5 x 8 x V oe (area of acceleration triangle, 4V)
(a) M1 12 x V oe (area of constant-velocity rectangle, 12V)
(a) M1 forms and simplifies 4V + 12V = 224 oe (16V = 224)
(a) A1 V = 14 cao
(a) Answer: V = 14 m/s
(b) M1 14 / 8 (ft from (a))
(b) A1 1.75 (m/s2) cao
(b) Answer: 1.75 m/s2
Question 10
B1 correct straight line from (0,0) rising to (15,12)
B1 correct horizontal line from (15,12) to (55,12)
B1 correct straight line falling from (55,12) to (65,0), then horizontal at 0 from (65,0) to (85,0)
B1 correct straight line rising from (85,0) to (97,12), with key time and velocity values labelled on the axes
Answer: Velocity-time graph with vertices at (0,0), (15,12), (55,12), (65,0), (85,0), (97,12).
Question 11
(a) M1 0.5 x 5 x 10 (= 25)
(a) M1 10 x 15 (= 150), and adds to the triangle area
(a) A1 175 (m) cao
(a) Answer: 175 m
(b) B1 160 (m)
(b) Answer: 160 m
(c) M1 175 - 160 (ft from (a) and (b))
(c) A1 Priya, by 15 (m) cao
(c) Answer: Priya, by 15 m
Question 12
(a) B1 B
(a) Answer: B
(b) B1 1.5 (s), ft from (a) if a straight line through zero was chosen
(b) B1 correct explanation, e.g. the highest point is where the velocity is zero, which is where the graph crosses the time axis
(b) Answer: t = 1.5 s, because this is where the velocity-time graph crosses the time axis (velocity = 0)
Question 13
(a) B1 awrt 8.4 (s), ft from a correct gradient calculation in (b)
(a) Answer: t = 8.4 s
(b) M1 0.5 x 4 x 6 + 6 x 2 (areas for 0-4s and 4-6s, = 12 + 12)
(b) M1 finds the time the velocity is zero within 6-10s (t = 8.4, ft from (a))
(b) M1 0.5 x 2.4 x 6 + 0.5 x 1.6 x 4 (splits the 6-10s section into positive and negative parts, = 7.2 + 3.2)
(b) M1 4 x 2 (area for 10-12s, = 8), and adds all magnitudes together
(b) A1 42.4 (m) cao
(b) Answer: 42.4 m
(c) M1 (12 + 12 + 7.2) - (3.2 + 8) (positive area minus negative area, ft from (b))
(c) A1 20 (m), in the original (positive/forward) direction cao
(c) Answer: 20 m (in the original direction of travel)
Question 14
(a) M1 identifies the trapezium rule structure: 0.5 x h x [(v0 + v5) + 2(v1 + v2 + v3 + v4)]
(a) M1 0 + 22 (= 22) and 2 x (8.5 + 15.0 + 19.0 + 21.0) (= 127)
(a) M1 0.5 x 1 x (22 + 127) oe
(a) A1 74.5 (m) cao
(a) Answer: 74.5 m
(b) B1 underestimate
(b) B1 correct reason, e.g. the velocity-time graph curves upward above the straight-line chords used by the trapezium rule (since the velocity increases at a decreasing rate), so the true area under the curve is greater than the trapezium estimate
(b) Answer: Underestimate, because the curve lies above the straight-line chords used by the trapezium rule.
Question 15
(a) M1 identifies gradient of the tangent = (24 - 4) / (4 - 0)
(a) M1 20 / 4
(a) A1 5 (m/s2) cao
(a) Answer: 5 m/s2
(b) B1 valid reason, e.g. the tangent is drawn/judged by eye, so its gradient only approximates the true gradient of the curve at that instant
(b) Answer: The tangent is drawn by eye, so its gradient is only an approximation of the curve's true gradient at t = 2 seconds.
Question 16
(a) M1 4 = (30 - u) / 5 oe
(a) M1 20 = 30 - u oe
(a) A1 u = 10 cao
(a) Answer: u = 10 m/s
(b) M1 0.5 x (10 + 30) x 5 oe (area of first phase using ft value of u, = 100)
(b) M1 30 x 20 (area of second phase, = 600)
(b) M1 forms equation 100 + 600 + 0.5 x t x 30 = 850 oe