A student drives a moped along a straight road as described by the velocity vers

Question

A student drives a moped along a straight road as described by the velocity versus time graph in the figure. The divisions along the horizontal axis represent 5.0 s and the divisions along the vertical axis represent 4.0 m/s. Sketch this graph in the middle of a sheet of graph paper. (Do this on paper. Your will need it to do part (a) and (b).) Directly above your graph, sketch a graph of the position versus time, aligning the time coordinates of the two graphs. (Do this on paper. Your instructor may ask you to turn in your work.) Sketch a graph of the acceleration versus time directly below the velocity-versus time graph, again aligning the time coordinates. On each graph, show the numerical values of x and ax for all points of inflection. (Do this on paper. Your instructor may ask you to turn in your work.) What is the acceleration at t = 30.0 s? m/s2 Find the position (relative to the starting point) at t = 30.0 s. m What is the moped's final position at t = 45.0 s? m A fly lands on one wall of a room. The lower-left corner of the wall is selected as the origin of a two-dimensional Cartesian coordinate system. The fly is located at the point having coordinates (2.05, 1.20) m. How far is it from the origin. m What is its location in polar coordinates? R = m theta = degree counterclockwise from the +x axis If the rectangular coordinates of a point are given by (5, y), and its polar coordinates are (r, 23degree), determine the value of y and r. y = r A surveyor measures the distance across a straight river by the following method. Starting directly across from a tree on the opposite bank, she walks d = 86 m along the riverbank to establish a baseline. Then she sights across to the tree. The angle from her baseline to the tree is d = 31.0degree. How wide is the river? m

Explanation / Answer

c) The acceleration is the slope of the line in the velocity/time graph. At t = 15.0 s, (six divisions from the origin) the slope is negative and is 4 vertical divisions divided by 2 horizontal divisions: 8/5 = 1.6 m/s² d) The position is ?v*dt; this is the total area under the graph between t = 0 and t = 15. This is area is made up of 3 sections from t = 0 to t = 7.5 the area is the area of a triangle of base 7.5 and altitude 8; the area is 0.5*7.5*8 = 30 from t = 7.5 to t = 12.5 the area is the area of the rectangle of base 5 and height 8 = 40 from t = 12.5 to t = 15, the area is the area of a triangle of base 2.5 and altitude 4 plus the rectangle of base 2.5 and height 4; the area is 0.5*2.5*4 + 2.5*4 = 15 add them up to get 30 + 40 + 15 = 85 m e) the first two parts of b) apply here. now there are two additional triangles to consider: one has a base of 5 and an altitude of 8 and the other a base of 5 and an altitude of -8. The areas are equal and opposite so cancel out, leaving the areas from the first two only = 70 m

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