D. Instantaneous velocity at time t=0.0 s, and at time t=1.0 sE. Average acceleration in the time interval t=0.0s to 1.0 sF. Instantaneous acceleration at time t=0.0 s, t=0.5 s, and t=1.0 s*I need a solution*Your answer should be the same as shown in the picture 7D. v (0 s) = 2mî, i (1 s) = - mî+ 1 mk7E. đav (0.0s 1.0 s) = -m/s î+ 1 m/s k7F. å (0 s) = (-4î- 2n²j+ nk) = (-4.00 î – 19.7j + 3.14k)á (1/2 s) = (-11+v2 (-) k)프= (-1.00 i + 1.35 k) 프, and%3D3%3D3тms2s2»m%3Ds2(-1.00 î + 1.35 k), and216.4à (1.0 s) =(-승1+2㎡"j-뜻시)프- (-0.444 1 + 19.7j - 2.47 R)을m(-0.444 î + 19.7 ĵ – 2.47 k)s2%3D9.4 7. The position coordinates of a particle are given by x = In(1+ 2t), y = 2 cos(nt),and z = tsin(nt/2) where x, y,and z are in meters while t is in seconds. Determine%3Dthe following quantities:Positicn vector at t= 0.0sand t- 1O

The position of the particle is given by, x=ln1+2 ty= 2 cos π tz=t sinπ t2

Answered: D. Instantaneous velocity at time t=0.0…

D. Instantaneous velocity at time t=0.0 s, and at time t=1.0 s E. Average acceleration in the time interval t=0.0s to 1.0 s F. Instantaneous acceleration at time t=0.0 s, t=0.5 s, and t=1.0 s

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter1: Units And Measurement

Section: Chapter Questions

Problem 81AP: Consider the equation s=s0+v0t+a0t2/2+j0t3/6+s0t4/24+ct5/120 , were s is a length and t is a time....

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Concept explainers

Units and Dimensions

These are generally defined as the terms which precede or describes the dimensions further. These are also further divided into two parts namely fundamental units and the derived units.

Question

D. Instantaneous velocity at time t=0.0 s, and at time t=1.0 s

E. Average acceleration in the time interval t=0.0s to 1.0 s

F. Instantaneous acceleration at time t=0.0 s, t=0.5 s, and t=1.0 s

*I need a solution

*Your answer should be the same as shown in the picture

7D. v (0 s) = 2mî, i (1 s) = - mî+ 1 mk
7E. đav (0.0s 1.0 s) = -m/s î+ 1 m/s k
7F. å (0 s) = (-4î- 2n²j+ nk) = (-4.00 î – 19.7j + 3.14k)
á (1/2 s) = (-11+v2 (-) k)프= (-1.00 i + 1.35 k) 프, and
%3D
3
%3D
3
т
m
s2
s2»
m
%3D
s2
(-1.00 î + 1.35 k), and
2
16.
4
à (1.0 s) =
(-승1+2㎡"j-뜻시)프- (-0.444 1 + 19.7j - 2.47 R)을
m
(-0.444 î + 19.7 ĵ – 2.47 k)
s2
%3D
9.
4

7. The position coordinates of a particle are given by x = In(1+ 2t), y = 2 cos(nt),
and z = tsin(nt/2) where x, y,and z are in meters while t is in seconds. Determine
%3D
the following quantities:
Positicn vector at t= 0.0sand t- 1O

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