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A travels along a straight road at a speed of 25 m/s while accelerating at 1.5 m

ID: 1817308 • Letter: A

Question

A travels along a straight road at a speed of 25 m/s while accelerating at 1.5 m/s2. At this now instant car C is traveling along the straight road with a speed of 30 m/s while decelerating at 3 m/s?. Determine the velocity and acceleration of car A relative to car C. (ans: v A/C = 21.5 m/s, theta v = 34.9 degree, a A/C = 4.20 m/s2, theta. = 75.4 degree) The man can row the boat in still water with speed of 5 m/s. If the river is flowing at 2 m/s. determine the speed of the boat and the angle theta he must direct the that it travels from A to B.

Explanation / Answer

Car B isn't part of the problem. Velocity is a vector. Car C is traveling down so velocity vector is down. Car A is traveling down and to the left (his right) at 45 deg. We can break his velocity up into a velocity in the same direction as C (which makes it look like A is going slower from people in C's perspective, and a velocity at right angles to C's motion (which add like sides of a right triangle). Sin45 = cos 45 = 0.707 25m/s*0.707 = 17.675m/s Relative to C A's velocity is SQRT(sideways velocity^2 + difference in up and down velocity^2) = SQRT[(17.675)^2 + (30-17.675)^2] = 21.548 m/s To find theta v look at the sides of the triangle arccos(17.675/21.548) = theta = 34.89 deg. Acceleration is the same story, it is a vector, so we can use vector addition. We break A's acceleration into two components, again following the 45 deg angle, but now C is acceleration towards the vertical component of A's acceleration, so they will add. 1.5m/s^2*0.707 = 1.0606601717798212866012665431573 m/s^2 Add the components using Pythagorean theorem again A's acceleration rel to C = SQRT[(1.06)^2 + (1.06+3)^2] = 4.197 m/s^2 Again the cosine of this angle is hypotenuse over adjacent so the angle is the inverse cos. arccos(1.06/4.197) = theta a = 75.36 deg. So, what did this problem test? Pythagorean theorem and adding vectors.It is always confusing which angles they are talking about, so the answers were a nice reassurance for the thetas.

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