An object is moving with uniform acceleration a. Its initial velocity

An object is moving with uniform acceleration a. Its initial velocity is u and after time t its velocity is v. The equation of its motion is v = u + at. The velocity (along y-axis) time (along x-axis) graph shall be a straight line

passing through origin
with x-intercept u
with y-intercept u
with slope u
This question was previously asked in
UPSC NDA-1 – 2018
The given equation of motion is v = u + at. We are asked about the velocity (v) vs. time (t) graph. This equation is in the form of a linear equation, y = mx + c, where v corresponds to y, t corresponds to x, ‘a’ (uniform acceleration) is the slope (m), and ‘u’ (initial velocity) is the y-intercept (c). The y-intercept is the value of y when x=0. In this case, it is the value of velocity (v) when time (t) is zero, which is the initial velocity ‘u’. Therefore, the graph is a straight line with a y-intercept equal to u.
– The equation v = u + at is a linear relationship between velocity (v) and time (t) for uniform acceleration.
– In a linear equation y = mx + c, ‘c’ is the y-intercept and ‘m’ is the slope.
– On a velocity-time graph, the y-axis represents velocity and the x-axis represents time.
The slope of the velocity-time graph represents the acceleration. Since the acceleration ‘a’ is uniform (constant), the graph is a straight line. The graph passes through the origin only if the initial velocity u is zero. The x-intercept would be the time when velocity is zero, which is t = -u/a (if applicable and physically meaningful).
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