The Position, Velocity, and Acceleration Vectors A motorist drives south at 20.0

Question

The Position, Velocity, and Acceleration Vectors A motorist drives south at 20.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.00 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.00-min trip, find (a) The total vector displacement, (b) The average speed, and (c) The average velocity. Let The positive x axis point east. When The Sun is directly overhead, a hawk dives toward The ground with a constant velocity of 5.00 m/s at 60.0 degree below The horizontal. Calculate The speed of its shadow on The level ground. Suppose The position vector for a particle is given as a function of time by (t) = x(i) + y(t) with x(t) = at + b and y(t) = ct2 + d, where a = 1.00 m/s, b = 1.00 m, c = 0.125 m/s2, and d = 1.00 m. (a) Calculate The average velocity during The time interval from t = 2.00 s to t = 4.00 s. (b) Determine The velocity and The speed at t= 2.00 s. The coordinates of an object moving in The xy plane vary with time according to The equations x = -5.00 sin (ot and y = 4.00 - 5.00 cos (ot, where (o is a constant, x and y are in meters, and t is in seconds, (a) Determine The components of velocity of The object at t = 0. (b) Determine The components of acceleration of The object at t = 0. (c) Write expressions for The position vector, The velocity vector, and The acceleration vector of The object at any time t > 0. (d) Describe The path of The object in an xy plot.

Explanation / Answer

There are multiple questions here. . i am allowed to answer only 1 at a time. I will answer question 1 for you.Please ask other as different question.

1.
In south , 20m/s for 30 min
displacement S1 = 20 m/s * (30*60)s = 36000 m (-j)

In west , 25m/s for 2 min
S2 = 25m/s * (2*60)s = 3000 m (-i)

In North west, 30m/s for 1 min
S3 = 30m/s * 60 s
= 1800 m [-cos45 i + sin 45 j]
= 1272.8 (-i) + 1272.8 (j)

a)
Total displacement,
S = S1+ S2 + S3
= -36000 j - 3000 i - 1272.8 i + 1272.8 j
= -4272.8 i -34727.2 j

magnitude= sqrt(4272.8^2 + 34727.2^2)
= 34989 m
direction, angle south of west = atan (34727.2/4272.8) = 83degree

b)
Total distance = 36000 + 3000 + 1800 = 40800 m
total time = 6min = 6*60 s = 360 s
average speed = total distance/totaltime = 40800/360 = 113.3 m/s

c)
Average velocity = total displacement /total time
= 34989/360
= 97.2 m/s

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