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Questions 1, 2, and 3 are short free-response questions that require about 13 minutes to answer and are worth 8 points. Questions 4 and 5 are long free-response questions that require about 25 minutes each to answer and are worth 13 points each. Show your work for each part in the space provided after that part.

The motion of an object is given by the following velocity-versus-time graph.(GRAPH IS NOT AVAILABLE TO COPY)(a) What is the displacement of the object from time $t=0$ s to $t=6 \mathrm{s}$ in the graph above?(b) At what times is the speed of the object increasing?(c) Make of a sketch of the object's position-versus-time graph during the time interval of $t=0$ s to $t=6$ s. Assume that the car begins at $x=0 .$

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Chapter 13

Practice Test 3

Section 2

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I think there is a velocity time curve as shown here in the figure. The time is represented over the X axis and velocity is represented over the y axis here, this is the origin. If the time is divided In two seconds like one, two, three, four, why, and six, the velocity is one m per second. The velocity is being measured in meters per second and two And in negative direction. Also this is -1 meter per second. No. Here we can do the partitions of the time axis which will be used in plotting the velocity in plotting their displacement time curb. Also, So here these are these partitions and The 6th to 2nd, the last second. And corresponding to this with all city time curb, we will plot the displacement time curve, Which is the 3rd part of the same problem. No, first of all, we plot this velocity time car in which the velocity is one m per second. Initially means when the time is zero, after which the velocity increases to two m/s, Up to a time off to seconds Here, this is two m per second. Now the velocity remains constant as two m per second For the further 1 2nd, then It starts reducing and becomes zero At five seconds. And then In the further 1 2nd The object moves in the opposite direction and the velocity becomes -1 m/s. Now we know the total displacement covered is given by the area covered by VT curb with time access. So displacement is given by the area cover between velocity time curb and the time access. And this is for the first part of the problem. So here the total displacement will be given by the area of the figure Up to two seconds, means this is even and this is a trapeze ium. After that area A two Up to three seconds, which is a rectangle, Then triangle up to five seconds, a three and finally when more triangles, A four up to last six seconds. Now for the area of trapeze um this is half into some of parallel sites and parallel sites are this one m per second and two m per second, so this is one plus two, multiplied by the gap between them, which is from 0 to 2 seconds. So this is two Plus A two area of rectangle, which is the height, which is two m per second, multiplied by the gap between them means three minus two seconds, which is one plus a three. The area of a triangle, right angled triangle, which is given by half into base into height, height again is two m per second and base is five minus three means this is too, and for a four as this is in the fourth quarter and so that will be negative. An area of triangle is having to based This is 1 6 -5 is one, And height is this -1. So Here it comes out to be three plus two plus two and minus 0.5 m. So finally, the total displacement made by this, shown by this vehicle is Given us 6.5 m comes out to be 6.5 m, which is the answer for the first part of the problem. Now, in the second part of the problem, we have to find the time at which the speed is increasing. So it is clear from the curb that speed is increasing initially from zero second to two seconds and then from five seconds to six seconds, but in the opposite direction. Now, in the third part of the problem, we have to convert this Whitaker into displacement time Kirk and foot. Which here we will use the same whitaker and corresponding to it, we are going to plot displacement time curve. Here they are. X axis is again divided into times one second, two seconds, three seconds, four seconds, five seconds and six seconds. And the distance over this, y axis is divided in two m which is one m, two m, three m, four m, five m, six m. Or if he compact the distances in order to accommodate all the distances, this is one, two, three for five, six and seven. Now in the first two seconds the distance covered was three m and that was For their distance from 0-2 seconds. The speed the velocity was increasing, so that will be the current one means it will be curved upward as the speed is increasing, the velocity is increasing like this. Now the velocity becomes constant and the distance covered in the next one second is further more two m. So that will be three plus two means hi meter initially this is three m, not the two m. So there is a connection here, so that is three m in two seconds and that is increasing velocity like this Now beyond two seconds Up to 32nd, The velocity was uniform. So the displacement time curve will be a straight line and extends further two m means it will be like this straight line, it should be a straight line now beyond it. The velocity starts decreasing and the distance covered. Yes, Further two m means up to seven m for five seconds, means of to help and the velocity is decreasing means it will be like this. They're decreasing curve whose one decreasing no in the next one second in the last one second, 30 stances 0.5 m and the direction is reversed means it will be 6.5 means here. So the displacement time curve is like this. So this is the overall transformation of viticulture into X. T. Curb here. This is X. And they're here. This is deep. Thank you.

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