A force-time (F-t) graph for a linear motion | vedclass

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(B) The change in linear momentum $(\Delta p)$ is equal to the area under the force-time $(F-t)$ graph.$\Delta p = \int_0^8 F \,dt = 	ext{Area under

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(B) The change in linear momentum $(\Delta p)$ is equal to the area under the force-time $(F-t)$ graph.$\Delta p = \int_0^8 F \,dt = 	ext{Area under the } F-t 	ext{ curve}$.From the graph,the area from $t=0$ to $t=2$ is a quarter-circle below the $t$-axis (negative area).The area from $t=2$ to $t=6$ is a semi-circle above the $t$-axis (positive area).The area from $t=6$ to $t=8$ is a quarter-circle below the $t$-axis (negative area).Given the radius of the circular segments is $r=2$ units (from $F=0$ to $F=2$ or $F=-2$):Area of quarter-circle $= \frac{1}{4} \pi r^2 = \frac{1}{4} \pi (2)^2 = \pi$.Area of semi-circle $= \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2)^2 = 2\pi$.Total area $= -(	ext{Area}_{0-2}) + (	ext{Area}_{2-6}) - (	ext{Area}_{6-8})$$\Delta p = -\pi + 2\pi - \pi = 0$.Thus,the linear momentum gained between $0$ and $8 \,s$ is $0$.

$A$ force-time $(F-t)$ graph for a linear motion is shown in the following figure. The segments shown are circular. The linear momentum gained between $0$ and $8 \,s$ is

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