The position-time graph of a body of mass 2, | vedclass

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(B) Mass of the body $m = 2\, kg$.From the graph,at $t = 0\, s$,the position $x = 0\, m$. The velocity $v_1$ is the slope of the graph at $t = 0\, s$,

Vedclass · Accepted Answer

$1.5$

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(B) Mass of the body $m = 2\, kg$.From the graph,at $t = 0\, s$,the position $x = 0\, m$. The velocity $v_1$ is the slope of the graph at $t = 0\, s$,which is $v_1 = 0\, m/s$.At $t = 4\, s$,the position $x = 3\, m$. For $t > 4\, s$,the graph is a horizontal line,meaning the velocity $v_2 = 0\, m/s$.Between $t = 0\, s$ and $t = 4\, s$,the body moves with a constant velocity $v = \frac{\Delta x}{\Delta t} = \frac{3 - 0}{4 - 0} = 0.75\, m/s$.Impulse at $t = 0\, s$: The body starts moving from rest,so the impulse is $J_1 = m(v_{initial} - v_{rest}) = 2(0.75 - 0) = 1.5\, Ns$.Impulse at $t = 4\, s$: The body comes to rest from a velocity of $0.75\, m/s$,so the impulse is $J_2 = m(v_{final} - v_{initial}) = 2(0 - 0.75) = -1.5\, Ns$.The magnitude of the impulse at both points is $1.5\, Ns$.

The position-time graph of a body of mass $2\, kg$ is as given in the figure. What is the impulse on the body at $t = 0\, s$ and $t = 4\, s$ (in $, Ns$)?

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