Which one of the following graphs represents the equation of motion v = u + at; where all quantities are non-zero and symbols carry their usual meanings ?
[amp_mcq option1=”(a)” option2=”(b)” option3=”(c)” option4=”(d)” correct=”option3″]
This question was previously asked in
UPSC NDA-2 – 2023
– v is the final velocity
– u is the initial velocity
– a is the acceleration (constant)
– t is the time
This equation is in the form of a linear equation y = mx + c, where v is the dependent variable (y), t is the independent variable (x), ‘a’ is the slope (m), and ‘u’ is the y-intercept (c). The problem states that all quantities (u, a, v, t) are non-zero, except t=0 at the start.
A graph of v versus t for this equation should be a straight line.
Since ‘u’ is non-zero, the line will intersect the v-axis (at t=0) at a point other than the origin.
Since ‘a’ is non-zero, the slope of the line will not be zero (the line is not horizontal).
Considering these points, graph (c) represents a straight line with a positive y-intercept (u > 0) and a positive slope (a > 0). While ‘a’ could be negative (negative slope) or ‘u’ could be negative (negative intercept), option (c) is the only graph that shows a straight line with a non-zero intercept and a non-zero slope, fitting the general form v = u + at with non-zero u and a. Graphs (a) and (d) are not straight lines, and graph (b) is a straight line passing through the origin, which would represent v = at (i.e., u=0), contrary to the condition that u is non-zero.
– A graph of velocity (v) versus time (t) for this equation is a straight line.
– The y-intercept of the line is the initial velocity (u).
– The slope of the line is the acceleration (a).