Suppose that the position vector for a particle is given as a function of time by (t) = x(t) + y(t), with x(t) = at + b and y(t) = ct2 + d, where a = 1.70 m/s, b = 1.20 m, c = 0.122 m/s2, and d = 1.18 m.
(a) Calculate the average velocity during the time interval from t = 2.05 s to t = 4.05 s.
= v (vector arrow on top)
(b) Determine the velocity at t = 2.05 s
= v (vector arrow on top)
Determine the speed at t = 2.05 s.
=m/s
PLEASE show your work. thank you very much
2 Answers

(a) The position at t = 2.05 s is
i (1.70*2.05 + 1.20) m + j (0.122*2.05^2 + 1.18) m
The position at t = 4.05 s is
i (1.70*4.05 + 1.20) m + j (0.122*4.05^2 + 1.18) m
The displacement during the time interval from 2.05 s to 4.05 s is
i (1.70 * 2) m + j (0.122) (2) (6.1) m = i 3.4 m + j 1.4884 m
The average velocity is displacement/time = 1.7 m/s i + 0.744 m/s j
(b) v = i [ d(at + b) / dt ] + j [ d(ct^2 + d) / dt ]
= ia + 2jct = 1.7 m/s i + 0.5002 m/s j
(c) The "speed" in (b) is
sqrt(1.7^2 + 0.5002^2) = use calculator, should be about 1.8 m/s

discover the area vector from the given equation, utilising the following substitutions x=t y = t²/4 r(t) = <t, t²/4> what's the fee of the parameter at (2,a million)? x = t t = 2 make sure with y(t) y = t²/4 = 2²/4 y = a million So, at (2,a million) the parameter "t" = 2 r'(t) = <a million, a million/2t> discover the unit tangent vector "T"; the cost vector is then 3T cm/s. T=r'(t)/r'(t) T= <a million, a million/2t>/?(a million+t²/4) Then for t=2 T(2) = <a million/?2, a million/?2> finally v = 3T v = <3/?2, 3/?2>cm/s