Q.If the displacement-time graph of a particle is represented by y = mx + c, then the particle is moving with (A) Constant speed
(B) Constant velocity
(C) Variable velocity
(D) Constant momentum

Differentiating x = 5t - t^2 gives v = 5 - 2t; substituting t = 2 s gives instantaneous speed = 1 m/s.

Given: x = 5t - t^2 (x in m, t in s).

Instantaneous velocity is defined as v = dx/dt.

Differentiating:

v = d/dt (5t - t^2) = 5 - 2t

At t = 2 s:

v = 5 - 2(2) = 5 - 4 = 1 m/s

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A linear $x$-$t$ graph $y=mx+c$ means constant velocity.

Velocity is the slope of the displacement-time graph, $v=dy/dx$ (treating $x$ as time here). For $y=mx+c$, the slope is $m$ everywhere — it never changes with time, so the particle moves with constant velo …

Instantaneous velocity is v = dx/dt; differentiating the given position equation and substituting t = 3 s gives 33 cm/s.

Given: x = at + bt^2, with a = 15 cm s^-1, b = 3 cm s^-2.

Instantaneous velocity: …

The slope of an x-t graph is $dx/dt$, which is exactly how velocity is defined.

For motion in a straight line, displacement $x$ is plotted against time $t$. Velocity is defined as the rate of change of displacement:

$v = \frac{dx}{dt}$

Graphically, $dx/dt$ at any point of the curve is the slope of the tangent to the $x$-$t$ graph at that point. For uniform motion the graph is a straight line and its constant slope gives the (constant) velocity; for non-uniform motion the slope changes from point to point and give …

Instantaneous velocity is the limiting value of average velocity as the time interval considered tends to zero; it equals dx/dt.

Average velocity over a time interval delta-t is defined as delta-x / delta-t, the displacement divided by the time taken. As delta-t is made smaller and smaller (delta-t -> 0), the average velocity approaches a fixed limiting value, which is called the instantaneous velocity at that instant:

v = lim(delta-t -> 0) (delta-x/delta-t) = dx/dt

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Differentiating x = a + bt^2 gives v = 2bt = 6t (since b = 3), which evaluates to 12 m/s at t = 2 s.

Given: x = a + b t^2, with a = 2 m, b = 3 m/s^2

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The slope of an x-t graph at any instant equals the particle's instantaneous velocity at that instant.

For a particle moving along a straight line, if x is its position at time t, the average velocity between two close instants t and t+Δt is v_avg = Δx/Δt, which is exactly the slope of the chord joining the two corresponding points on the x-t graph. As Δt is made smaller and smaller (Δt → 0), the chord becomes the tangent to the curve at that point, and the average velocity becomes the instantaneous velocity: v = dx/dt. So at every point on an x-t graph, the slope of the tangent drawn at that point gives the instantaneous velocity of the particle at that instant …