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Q.An object is moving with a constant acceleration. Draw time-velocity and time-displacement graphs for the object.

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2020Subjective· 2mImportance★★★★★
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Constant acceleration gives v=u+atv = u+at (a straight line in vv-tt) and s=ut+12at2s = ut+\tfrac12 at^2 (a parabola in ss-tt).

Step 1 — Velocity-time graph.

For constant acceleration aa and initial velocity uu,

v=u+atv = u + at

This is a linear (straight-line) equation in tt: intercept on the vv-axis is uu, and the slope of the line equals the acceleration aa. If a>0a>0, the line rises steadily; if a<0a<0 (deceleration), it falls steadily. The area under this line up to time tt equals the displacement travelled.

Step 2 — Displacement-time graph.

Integrating v=u+atv=u+at with respect to time,

s=ut+12at2s = ut + \frac{1}{2}at^2

This is a quadratic equation in tt, so the ss-tt graph is a parabola. Starting from the origin (assuming s=0s=0 at t=0t=0), the curve bends increasingly upward over time if a>0a>0 (the slope of this curve at any instant is the velocity, which itself is increasing) — i.e., it is concave up. If a<0a<0, the curve bends over (concave down) as the body decelerates.

Step 3 — Sketch description. …

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