Kinematics graphs Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Kinematics graphs questions. See exactly how to solve problems on v-t graph, area under graph, displacement, rectangle area.

v-t grapharea under graphdisplacementrectangle areaconstant velocitygradient
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A velocity-time graph for a particle is a single straight line rising from the origin (0,0)(0,0) to the point (4,12)(4,12), where time tt is in seconds and velocity vv is in m/s. Find the displacement of the particle during the first 44 seconds.

Worked solution

  1. Identify the region under the velocity-time graph

    triangle: base 4 s, height 12 m/s\text{triangle: base } 4\text{ s, height } 12\text{ m/s}

    Displacement is the area between the line and the time-axis.

  2. Apply the area formula for a triangle

    s=12×4×12s = \tfrac12 \times 4 \times 12

    Area of a triangle is half the base times the height.

  3. Evaluate the displacement

    s=24 ms = 24\text{ m}

    This is the displacement over the interval.

Answer
24 m24\text{ m}
Question 2
2 markseasy
A velocity-time graph for a particle is a single straight line rising from the origin (0,0)(0,0) to the point (6,10)(6,10), where time tt is in seconds and velocity vv is in m/s. Find the displacement of the particle during the first 66 seconds.

Worked solution

  1. Identify the region under the velocity-time graph

    triangle: base 6 s, height 10 m/s\text{triangle: base } 6\text{ s, height } 10\text{ m/s}

    Displacement is the area between the line and the time-axis.

  2. Apply the area formula for a triangle

    s=12×6×10s = \tfrac12 \times 6 \times 10

    Area of a triangle is half the base times the height.

  3. Evaluate the displacement

    s=30 ms = 30\text{ m}

    This is the displacement over the interval.

Answer
30 m30\text{ m}
Question 3
2 markseasy
A velocity-time graph for a particle is a single straight line rising from the origin (0,0)(0,0) to the point (5,8)(5,8), where time tt is in seconds and velocity vv is in m/s. Find the displacement of the particle during the first 55 seconds.

Worked solution

  1. Identify the region under the velocity-time graph

    triangle: base 5 s, height 8 m/s\text{triangle: base } 5\text{ s, height } 8\text{ m/s}

    Displacement is the area between the line and the time-axis.

  2. Apply the area formula for a triangle

    s=12×5×8s = \tfrac12 \times 5 \times 8

    Area of a triangle is half the base times the height.

  3. Evaluate the displacement

    s=20 ms = 20\text{ m}

    This is the displacement over the interval.

Answer
20 m20\text{ m}
Question 4
2 markseasy
A velocity-time graph for a particle is a single straight line rising from the origin (0,0)(0,0) to the point (3,20)(3,20), where time tt is in seconds and velocity vv is in m/s. Find the displacement of the particle during the first 33 seconds.

Worked solution

  1. Identify the region under the velocity-time graph

    triangle: base 3 s, height 20 m/s\text{triangle: base } 3\text{ s, height } 20\text{ m/s}

    Displacement is the area between the line and the time-axis.

  2. Apply the area formula for a triangle

    s=12×3×20s = \tfrac12 \times 3 \times 20

    Area of a triangle is half the base times the height.

  3. Evaluate the displacement

    s=30 ms = 30\text{ m}

    This is the displacement over the interval.

Answer
30 m30\text{ m}
Question 5
2 markseasy
A particle moves at a constant velocity of 1212 m/s. On a velocity-time graph this is a horizontal line from (0,12)(0,12) to (5,12)(5,12). Find the displacement of the particle between t=0t=0 s and t=5t=5 s.

Worked solution

  1. Recognise constant velocity as a horizontal line

    v=12 m/s (constant)v = 12\text{ m/s (constant)}

    A flat v-t line means the velocity does not change.

  2. The area under the line is a rectangle

    s=12×(5−0)s = 12 \times (5-0)

    Displacement equals width in time multiplied by the height (velocity).

  3. Evaluate the displacement

    s=60 ms = 60\text{ m}

    The rectangle area gives the displacement.

Answer
60 m60\text{ m}

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