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
- A.passing through origin
- B.with x-intercept u
- C.with y-intercept u
- D.with slope u
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Explanation
The correct answer is C, with y-intercept u. The equation v = u + at has the form y = c + mx, so the graph is a straight line that cuts the velocity axis at u and has slope a.
On a velocity-time graph, time is on the x-axis and velocity on the y-axis. At t = 0 the velocity is u, so the line starts from the point (0, u) on the y-axis; that value is the y-intercept. The slope of a velocity-time graph gives the acceleration, here a, and the area under the line gives the displacement. Only if the body starts from rest (u = 0) does the line pass through the origin.
A is wrong: the line passes through the origin only when u = 0, which is not given.
B is wrong: u is the value of velocity, read on the y-axis, not an x-intercept.
D is wrong: the slope of the line is the acceleration a, not the initial velocity u.
Exam tip: on a v-t graph, slope = acceleration, area under the graph = displacement, y-intercept = initial velocity.