CDS (I) 2023 General Knowledge · Question 9 of 120

Correct — B, its net displacement. On a velocity-time graph the vertical axis carries velocity and the horizontal axis carries time, so any small strip of area under the line is a velocity multiplied by a time interval — which is a length. Adding the strips over the whole interval adds up all those small lengths, and the total is how far the particle has moved. The area is a signed quantity: while the velocity is positive the strip lies above the time axis and counts as positive, and if the velocity turns negative the strip lies below the axis and subtracts. A signed total of that kind is displacement, not distance, which is exactly why the key names net displacement.

• (a)its average velocity — Average velocity is displacement divided by the time taken, so it is the area divided by the width of the interval, not the area itself. The units alone rule it out — an area on this graph is measured in metres, an average velocity in metres per second.
• (c)the distance travelled by it — True only in the special case where the velocity never changes sign. If the particle decelerates, stops and comes back, part of the area lies below the time axis; distance would add that part, whereas the area subtracts it. The area gives the net figure, which is displacement.
• (d)its average speed — Fails on the same two counts as option (a). It is a rate rather than a length, and it is built from distance rather than from a signed total, so it cannot be what an area under this graph gives.

Motion graphs turn calculus into geometry. On a distance-time graph the gradient is speed. On a velocity-time graph the gradient is acceleration, and the area between the line and the time axis is displacement. Uniform acceleration makes the line straight, so the area is a simple trapezium and the standard equations of motion drop straight out of it — the trapezium's area, half the sum of the initial and final velocities multiplied by the time, is precisely s = ut + at squared over two.

Options (b) and (c) are the pair that has to be separated, and the separation is the difference between a signed and an unsigned total. For a particle that only ever moves forward the two coincide, and a student who has only met that case may well feel that (c) is just as good an answer. It is not, because area below the time axis is negative area, and displacement is the quantity that respects the sign. The paper's key names net displacement, and the word 'net' is the giveaway that signs are meant to matter. Options (a) and (d) fail more cheaply still: both are rates, and no area under a velocity-time graph can be a rate.

• On a velocity-time graph the gradient gives acceleration and the area under the line gives displacement.
• The area is signed — velocity below the time axis contributes negative area — so it yields displacement rather than distance.
• Distance and the magnitude of displacement agree only when the motion never reverses direction.
• Uniform acceleration gives a straight line, whose trapezium area reproduces s = ut + at^2/2.
• On a distance-time graph, by contrast, it is the gradient and not the area that carries the physical meaning.

• Treating distance and displacement as interchangeable because they usually agree in textbook examples.
• Using the area of a velocity-time graph to find average velocity without dividing by the time.
• Carrying over the habit from a distance-time graph, where it is the gradient rather than the area that matters.

Either as this conceptual item on what the area represents, or as a numerical where the area of a trapezium under the line has to be computed to get the displacement.

• practice — not a real PYQ

  The slope of a velocity-time graph for a body moving in a straight line gives its

  • (a)displacement
  • (b)acceleration
  • (c)average speed
  • (d)momentum

  Answer(b) acceleration — the gradient is a change in velocity divided by the time taken, which is the definition of acceleration.

• practice — not a real PYQ

  A body starts from rest and accelerates uniformly to 20 m/s in 10 seconds. The displacement in that interval, read as the area under its velocity-time graph, is

  • (a)50 m
  • (b)100 m
  • (c)200 m
  • (d)400 m

  Answer(b) 100 m — the graph is a triangle of base 10 s and height 20 m/s, so the area is half of 10 times 20.