1. The vector position of a particle varies in time according to the expression

Question

1. The vector position of a particle varies in time according to the expression r = (9.00 i -9.40t^2 j) m. What is the velocity and acceleration as a function of time?

2. For the particle in the question 1 above , Determine the particle's position an velocity at t= 1.00s.

3. A particle initially located at the origin has an acceleration of a=2.00 j m/s^2 and an initial velocity of v0=6.00 i m/s. Write the expression for vectors position and the velocity at any time t.

4. For the particle in the question 3 abovre, Find the position and velocity at t=8.00s.

5. An athlete rotates a 1.00 kg discus along a circular path of radius 3.281 ft. The maximum speed of the discus is 17.0 m/s. Determine the magnitude of the maximum radial acceleration of the discus.

PLEASE SHOW ALL STEPS USED TO GET FINAL ANSWER!!!!

Explanation / Answer

1. r = (9.00 i -9.40t^2 j) m
v = dr/dt = -18.8t j m/s
a = dv/dt = -18.8 j m/s^2

2. for T=1s

v = dr/dt = -18.8j m/s
a = dv/dt = -18.8 j m/s^2

3.

a .

v = u + a dt
v = 6i + 2t j
the vector position ;
r = ro + v dt
= 0 + 6t i + t^2 j m
= 6t i + t^2 j m

b .

(b)v = 6i + 2t j m/s

4.

(d) v(8) = 6i+ 16j
l vl = sqrt ( 6^2 + 6^2) = 8.485 m/s

5. a = v^2/r = 289/1.00 = 289ms^-2

(also F = ma = mv^2/r = 1*289/1.00 = 289Kgms^-2)

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