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Jusaquikie
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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
Jusaquikie Group Title
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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Jusaquikie Group TitleBest 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

Aperogalics Group TitleBest 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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