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An iceboat has a constant velocity toward the east when a sudden gust of wind causes the iceboat to have a constant acceleration toward the east for a period of 3.0 s. A plot of x versus t is shown in the figure below, where t = 0 is taken to be the instant the wind starts to blow and the positive x axis is toward the east.
(a) What is the acceleration of the iceboat during the 3.0 s interval?
(b) What is the velocity of the iceboat at the end of the 3.0 s interval?
(c) If the acceleration remains constant for an additional 3.0 s, how far does the iceboat travel during this second 3.0 s int
 one year ago
 one year ago
An iceboat has a constant velocity toward the east when a sudden gust of wind causes the iceboat to have a constant acceleration toward the east for a period of 3.0 s. A plot of x versus t is shown in the figure below, where t = 0 is taken to be the instant the wind starts to blow and the positive x axis is toward the east. (a) What is the acceleration of the iceboat during the 3.0 s interval? (b) What is the velocity of the iceboat at the end of the 3.0 s interval? (c) If the acceleration remains constant for an additional 3.0 s, how far does the iceboat travel during this second 3.0 s int
 one year ago
 one year ago

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JusaquikieBest ResponseYou've already chosen the best response.0
I just don't think i have enough information to work this problem please help me see where i need to start.
 one year ago

AperogalicsBest ResponseYou've already chosen the best response.1
(b) for the velocity at the end of 3sec just take the area under the graph from x=0 to x=3 so here just simply count the area blocks that comes out to be approx. 12+something (a) for that u have to calculate the change in x(distance) at given interval of time i.e. 00.5,0.51,.......so on and then plot the graph between change in x/0.5(i.e v) versus T then calculate area under it u will get acceleration. (c) for it use second law of motion i.e s=ut+1/2*at^2:)
 one year ago
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