Question

Suppose the position vector for a particle is given as a function of time by r (t) = x(t)î + y(t)ĵ, with x(t) = at + b and y(t) = ct2 + d, where a = 1.60 m/s, b = 1.05 m, c = 0.127 m/s2, and d = 1.20 m.

(a) Calculate the average velocity during the time interval from t = 2.10 s to t = 4.25 s.

(b) Determine the velocity at t = 2.10 s.

dDetermine the speed at t = 2.10 s

Please show work

Answer 1

Position vector is given by
= x i + y j
=[ (at +b) i + (ct²+d) j ]
Now putting all the values of a, b, c and d
= [(1.6t + 1.05) i + (0.127t² +1.2) j]
(a) the distance at t = 2.1 s
(2.1)= [(1.6*2.1 + 1.05) i +(0.127*(2.1)² +1.2) j]
= [4.41 i + 1.76 j ]
Now the position vector at t = 4.25
(4.25) = [(1.6*4.25 + 1.05) i +(0.127*(4.25)² +1.2) j]
= 7.85 i + 3.494 j
now we know that the average velocity is given by
V= total displacement /total time = [(4.25) - (2.1)] /(4.25-2.1 )
V = (3.44 i + 1.734 j) /(2.15)
V = 1.6 i + 0.806 j
(b)
Now we know that velocity is given by
= d/dt
= 1.6 i + (0.127*2t) j
= 1.6 i + 0.254*t j
Now the velocity at t = 2.1 s
(2.1) = 1.6 i +(0.254*2.1) j
(2.1) = 1.6 i + 0.5334 j
(d) Now speed at t = 2.1 s
V = (1.6² + 0.5334²)^1/2 = 1.686 m/s