A parachutist bails out and freely falls 50 m. Then the parachute opens, and thercafter she dec

A parachutist bails out and freely falls 50 m. Then the...

Key Concepts

Motion under Gravity

This concept involves the study of objects moving under the influence of gravity, typically near Earth’s surface where gravitational acceleration is approximately constant. It is fundamental for analyzing free-fall scenarios where no other forces significantly affect the motion except gravity.

Uniformly Accelerated Motion

Uniformly accelerated motion refers to movement in which the acceleration is constant. This principle is central to kinematics and is used to model both the free-fall phase (with constant gravitational acceleration) and the deceleration phase (with a specified constant deceleration) in various problems.

Kinematic Equations

Kinematic equations provide mathematical relationships between displacement, velocity, acceleration, and time for objects undergoing uniformly accelerated motion. These equations are instrumental in solving problems that involve determining unknown variables when certain other variables are known.

Piecewise Motion Analysis

This concept involves breaking down a complex motion into distinct segments, each with its own constant acceleration. By analyzing each segment separately using appropriate equations, one can combine the individual results to understand the overall motion, such as calculating total time of flight or total displacement.

Transcript

00:01 In this problem of line integrals, we have to find the work done by force field we have given f of xy is equals to xi plus yj by moving counterclockwise along the given path i'm showing here.

00:20 So this is the given path and we are moving in anti -clockwise direction or we can say counterclockwise direction from the point 0 to 1 .10.

00:30 And then 01 from 1 0 to 0 1 and then 0 1 to 0 0 0 0.

00:36 So work done is given by here integration of f.

00:41 .d .d .r along the curve is a piecewise function so we can name it as c1, this is c2 and c3.

00:49 So this would be equal to say work done would be equal to integration of say f .dddd along the curve c1.

00:57 Then integration of f dot dr along the curve c1.

00:59 D r along the curve c2 and integration of f dot d r along the curve c3 so for this first we have to find the value of r this is r t so curve c is such that r t is equal to first we are moving from 0 to 10 that means x is varying from 0 to 1 and y is equal to 0 so this would be t i such that t is the value from 0 to 1 now we are moving from one 0 to 0 1 so we can parameterize it as 2 minus t i plus t minus 1 j where value of t is varying from 1 to 2 and from 0 1 to 0 here y is equal to 1 and here y is equal to 0 and x is 0 at this point and this point also so we can parameterize at 3 minus t where value of t would be varying from 2 to 3.

02:09 Now for curve c1 so we can write it as for curve c1.

02:18 This is r t is equal to t i so we have r t is equal to t i.

02:24 Now when we differentiate it so this value would be d r is equal to simply i and multiplied with d t i d t now here x is equal to t and y is equal to 0 for curve c1 now for curve c2 for curve c2 r t is equal to r t is equal to 2 minus t i so this is 2 minus t i plus t minus 1 j and value of t is here from 1 to 2 now we have to find the value of d r so d r would be equal to this is minus i and plus j and multiplied with d t now for curve c3 for curve c3 r t is equals to 3 minus t j that means d r would be equals to minus j and multiplied with d t and also we can say here for curve 2 x is equals to 2 minus t and y is equals to t minus 1 for curve c 3 x is equal to 0 and y is equal to 3 minus t now we have given f of x y is x i plus y j so for this first we would find f dot d r for curve c1 say for curve c1 this is w is equal to and t is varying from 0 to 1 so we have to put the limit of t which is 0 to 1 f is equals to here x is equal to t so this value would be t i and multiplied with simply i d t so this is equal to t d t now plus t d t plus this is plus now integration f dot d r along the curve c2 and value of t is varying from 1 to 2 x is equal to 2 minus t and y is equal to t minus 1 now when we multiply with it dot product so this is equal to here this value would be t minus 2 plus t minus 1 and d t now for the curve c3, we have x is equal to 0 and y is equal to 3 minus t multiplied with this is 3 minus t j multiplied with minus j so this is equals to we can say this is t minus 3 and here this is d t and we have the integration limits of t which is varying from 2 to 3 now we have to integrate it and then add it so w is equal to integration of t is simply t squared with 2 and integration with limit for from 0 to 1.

05:19 Integration of this would be t plus t that is 2 t minus 3.

05:25 So this value would be here integration t square minus 3 t and limits are from 1 to 2.

05:33 Now integration of t minus 3 is simply t squared with 2.

05:37 So this is t squared divided with 2 minus 3 t and limits from 2 to 3...

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