A body in one-dimensional motion has zero spe | vedclass

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(A) The magnitude of velocity is defined as speed.Mathematically,$v = |\vec{v}|$,where $v$ is speed and $\vec{v}$ is velocity.If the speed $v = 0$ at

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Zero velocity

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(A) The magnitude of velocity is defined as speed.Mathematically,$v = |\vec{v}|$,where $v$ is speed and $\vec{v}$ is velocity.If the speed $v = 0$ at an instant,then the magnitude of the velocity must also be zero.Therefore,the velocity $\vec{v}$ must be zero at that instant.Note that the acceleration does not necessarily have to be zero; for example,a ball thrown vertically upward has zero velocity and zero speed at its highest point,but it still experiences a constant acceleration due to gravity $(g = 9.8 \ m/s^2)$.

Column-$I$	Column-$II$
$(1)$ Acceleration is positive	$(a)$ Speed of the particle decreases
$(2)$ Acceleration is negative	$(b)$ Speed of the particle increases
	$(c)$ Speed of the particle keeps changing

$A$ body in one-dimensional motion has zero speed at an instant. At that instant,it must have:

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The displacement of a particle,moving in a straight line,is given by $s = 2t^2 + 2t + 4$,where $s$ is in metres and $t$ is in seconds. The acceleration of the particle is ........ $m/s^2$.

Motion of a particle is given by the equation $s = (3t^3 + 7t^2 + 3t + 8) \ m$. The acceleration of the particle at $t = 1 \ s$ is: (in $m/s^2$)

Match the items in Column-$I$ with the items in Column-$II$ correctly.

Column-$I$ Column-$II$

$(1)$ Acceleration is positive $(a)$ Speed of the particle decreases

$(2)$ Acceleration is negative $(b)$ Speed of the particle increases

$(c)$ Speed of the particle keeps changing

If $v = x^2 - 5x + 4$,find the acceleration of the particle when the velocity of the particle is zero.

Which physical quantity can be found by first differentiation and second differentiation of position vector?

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