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Q.A particle is moving along X-axis with uniform positive acceleration.

(a) Obtain an expression for displacement by drawing v-t graph.
(2)
(b) A ball is thrown vertically upwards with a velocity of 20 m/s from the top of a tower of height 25 m from ground. How long does it remain in air ? (g = 10 ms⁻²) (2)
Kerala DhseKerala DHSE Plus One Board 2022Subjective· 4mImportance★★★★★
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Only the general stem "a particle moves along the X-axis with uniform positive acceleration" was extracted, without the specific sub-parts/numbers this 4-mark question likely asked for (such as given values of u, a, or a specific graph to sketch/read) — so this answer honestly covers the general theory and graph shapes for this kind of motion, rather than a numerical worked example that isn't actually available from the given data.

For a particle moving along the X-axis with a uniform (constant) positive acceleration a, starting with initial velocity u at t = 0:

  1. Velocity–time relation: v = u + at

    This is a straight line on the v–t graph with slope equal to a (positive), and v-intercept u.

  2. Position–time relation: x = ut + ½at²

    This is a quadratic in t, so the x–t graph is a parabola opening upward, since a > 0. Its slope (which represents velocity) keeps increasing with time — the curve gets steeper and steeper, unlike the straight-line x–t graph of a body moving at constant velocity.

  3. Velocity–position relation: v² = u² + 2ax (or v² = u² + 2a·s for displacement s)

  4. Acceleration–time relation: since the acceleration is uniform (constant) and positive, the a–t graph is simply a horizontal straight line at height a, for all values of t.

If, additionally, the particle starts from rest (u = 0), then v = at and x = ½at² — the v–t line passes through the origin, and the x–t parabola starts flat at the origin and curves upward increasingly steeply.

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