The displacement (x) - time (t) graph of a pa | vedclass

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(A) The slope of the displacement-time graph represents the velocity of the particle,i.e.,$v = \frac{dx}{dt}$.At $t = 0$,the graph starts from the ori

Vedclass · Accepted Answer

Particle starts with zero velocity and variable acceleration

Vedclass · Answer

(A) The slope of the displacement-time graph represents the velocity of the particle,i.e.,$v = \frac{dx}{dt}$.At $t = 0$,the graph starts from the origin $(0, 0)$ and the tangent to the curve at the origin is horizontal (parallel to the $t$-axis).Since the slope of the tangent at $t = 0$ is zero,the initial velocity of the particle is zero.As time increases,the slope of the curve changes continuously,which implies that the velocity is changing with time.Since the velocity is changing with time,the particle is undergoing acceleration.Because the slope (velocity) is not changing at a constant rate (the curvature of the graph is not parabolic),the acceleration is variable.Therefore,the particle starts with zero velocity and variable acceleration.

$(1)$ Velocity decreases	$(a)$ Linear graph (constant slope)
$(2)$ Velocity increases	$(b)$ Concave up graph (increasing slope)
$(3)$ Velocity is constant	$(c)$ Concave down graph (decreasing slope)

The displacement $(x)$ - time $(t)$ graph of a particle is shown in the figure. Which of the following is correct?

Similar Questions

Match Column-$I$ with Column-$II$ based on the distance-time $(d-t)$ graphs provided in the image.
$(1)$ Velocity decreases $(a)$ Linear graph (constant slope)
$(2)$ Velocity increases $(b)$ Concave up graph (increasing slope)
$(3)$ Velocity is constant $(c)$ Concave down graph (decreasing slope)

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From the given $v-t$ graph,find the ratio of distance to displacement in $25\,s$ of motion.

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