Cars A and B move in the same direction in adjacent lanes. The position x of car

Question

Cars A and B move in the same direction in adjacent lanes. The position x of car A is given in the figure, from time t = 0 to t = 7.0 s. At t = 0, car B is at x = 0, with a velocity of 4.0 m/s and a negative constant acceleration aB. (a) What must aB be such that the cars are side by side (momentarily at the same value of x) at t = 20. s? m/s2 (b) For that value of aB, how many times are the cars side by side? times (c) Sketch the position x of car B versus time t. (Do this on paper. Your instructor may ask you to turn in this sketch.) (d) How many times will the cars be side by side if the magnitude of acceleration aB is more than the answer to part (a)? times (e) How many times will the cars be side by side if the magnitude of acceleration aB is less than the answer to part (a)? times

Explanation / Answer

So you have this equation: x-x0 = v0t + 1/2at^2 x0=initial position vo= initial velocity For car B, you know that v0=11.8 m/s and they ask at t=4.10 sec., so now you need to know the x0 value, which it states in the question that "the cars are momentarily at the same value of x at t=4", so according to the picture, you would find x Wait, are you sure that's the right picture?? The picture shows a velocity graph, however in the problem it says that "The POSITION of car A is given from time t=0 to t=7.0 s". and the picture only shows it up to 6 seconds.

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