NDA (I) 2018 General Ability Test · Question 54 of 150

Correct — C, with y-intercept u. Lay the equation of motion beside the equation of a straight line. The line is y = mx + c, with m the slope and c the intercept on the y-axis. The equation given is v = at + u, and the question puts velocity on the y-axis and time on the x-axis. Matching term for term, v plays the part of y, t plays the part of x, the acceleration a is the slope, and the initial velocity u is the constant left over — which is the y-intercept. Set t to zero and the graph gives v equal to u, exactly as it should, because u is the velocity the object already had when the clock started.

• (a)passing through origin — That would require the velocity to be zero when the time is zero, which means u equal to zero. The question says the initial velocity is u, a quantity it has taken the trouble to name, so the line starts above the origin. A line through the origin describes an object starting from rest.
• (b)with x-intercept u — The intercept on the time axis is where the velocity falls to zero, which happens at a time of minus u divided by a, not at u. Beyond that, u is a velocity and the x-axis carries time, so the two cannot even share a unit.
• (d)with slope u — The slope of a velocity-time graph is the rate at which velocity changes, which is the acceleration a. Putting u there confuses the starting value with the rate of change — the classic mix-up between the intercept and the slope.

For motion with uniform acceleration the three equations of motion are v = u + at, s = ut + half a t squared, and v squared = u squared plus 2as. The first of these is linear in time, so plotting velocity against time gives a straight line whose slope is the acceleration and whose intercept on the velocity axis is the initial velocity. The area under that line between two instants gives the displacement over that interval.

This item is really a mathematics question wearing a physics costume, and the reliable method is to force the physical equation into the shape y = mx + c before answering anything. Rewriting v = u + at as v = (a)t + (u) does the whole job in one step and shows immediately that a is the slope and u the intercept. Two habits follow from it. First, always check which quantity is on which axis, because the same equation drawn with time on the vertical axis would give a different answer. Second, run a units check on any option that offers to put a velocity on a time axis — the mismatch alone kills option (b).

• On a velocity-time graph the slope gives the acceleration and the area under the curve gives the displacement.
• In v = u + at the initial velocity u is the intercept on the velocity axis and the acceleration a is the slope.
• A velocity-time graph through the origin means the body started from rest.
• For a body slowing down the acceleration is negative, so the line slopes downwards and meets the time axis when the body stops.
• On a displacement-time graph, by contrast, the slope gives the velocity.

• Swapping the slope and the intercept, and so choosing u as the slope.
• Assuming every motion graph passes through the origin.
• Not noticing which quantity the question has put on which axis.

NDA asks this either in words, as here, or by printing four candidate graphs and asking which one represents v = u + at.

• practice — not a real PYQ

  On a velocity-time graph for uniformly accelerated motion, the area under the line between two instants gives the

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

  Answer(b) displacement — area on a velocity-time graph is velocity multiplied by time, which is displacement.

• practice — not a real PYQ

  A body starts from rest and accelerates uniformly. Its velocity-time graph is a straight line

  • (a)parallel to the time axis
  • (b)passing through the origin
  • (c)with a negative slope
  • (d)with a non-zero intercept on the velocity axis

  Answer(b) passing through the origin — starting from rest means u is zero, so the intercept vanishes.