Part A: A particle moves in the xy plane with constant acceleration. At time t=0

Question

Part A: A particle moves in the xy plane with constant acceleration. At time t=0 s, the position vector for the particle is r=6.10mx +8.50my. The acceleration is given by the vector a=5.30m/s2x +4.20m/s2y. The velocity vector at time t=0 s is v=9.70m/sx - 5.60 m/sy. Find the magnitude of the velocity vector at time t = 6.20 s.

Part B: What is the angle between the velocity vector and the positive x-axis at time t=6.20s?

Part C: What is the magnitude of the position vector at time t=6.20s?

Acceleration, Velocity. and Displacement Vector Part A 6.1 mz + 8.50my. The acceleration is given by the vector a 5.30 822 + 4 m 2 The A particle moves in the a plans with constant acceleration. At time t-0 s the position vector or the perticle is r T- 9.7Uhrasi-5.60asy. Find the magnltude ot the velacity vector at timat 6.20 s. 41.2 m/s eboty vector at t e t-o s is ara corract. Part R what is the angle between the velocity vector and the positivex-axis at timct .20 a? 51 deg Submit Answer linea-ties 2/ Part C what the magnitude of the pusition vector dt tirre t-5.2U s? Submit Answer Tr 0/0 Part U Suhmt AnTres 1998-2015 by Honde State University. All rights reserved The contents cfthis web sita are or use by the students enrolled in this course only. Any Cistribution cfthe contents n whole or in parts and in any form, nduding posting on the web cr pr nted er en s*,riclation cf the Cop n ht laws and the FSU honor codes his discussion is closed. b, send Feedbace

Explanation / Answer

x-component of Position , velocity and acceleration at t= 0 are 6.1 , 9.7 and 5.3 respectively.
using v = u + at ,in x- direction, we get
Vx(6.2 sec.) = 9.7 + 5.3*6.2 = 42.6 m/s
Using s = r(f) - r(i) = ut +at2/2 in x-direction, we get
x(6.2) - 6.1 = 9.7*6.2 + 5.3*6.22 / 2
x(6.2) = 168.1 m
Similarly working y- direction we get,
Vy(6.2) = -5.6 + 4.2*6.2 = 20.4 m/s
y(6.2) - 8.5 = -5.6*6.2 + 4.2*6.22/2
y(6.2) = 54.5 m

A) magnitude of velocity V = ( Vx2 + Vy2 )1/2
                                           = ( 20.42 +42.62)1/2 = 47.2 m/s

B) Angle made by velocity with +ve x-axis is given by
theta = tan-1(Vy/Vx) = tan-1(20.4/42.6) = 25.6 deg

C) Magnitude of position vector,
r = (x2 +y2 )1/2 = (168.12 + 54.52) 1/2 = 176.7 m

D) Angle between position vector and x-axis
theta = tan-1(y/x) = tan-1(54.5/168.1) = 18 deg

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