Answer to Question #94823 in Mechanics | Relativity for stefanie

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 = 2.00 m/s, b = 1.40 m, c = 0.129 m/s2, and d = 1.12 m.
(a) Calculate the average velocity during the time interval from t = 2.20 s to t = 4.05 s.
v
= m/s

(b) Determine the velocity at t = 2.20 s.
v
= m/s

Determine the speed at t = 2.20 s.
m/s

Answer 1

The position vector for a particle is given by equation

${\bf r}(t)=x(t){\bf\hat i}+y(t){\bf\hat j}$

${\bf r}(t)=(at+b){\bf\hat i}+(ct^2+d){\bf\hat j}$

${\bf r}(t)=(2.00t+1.40){\bf\hat i}+(0.129t^2+1.12){\bf\hat j}$

(a) The average velocity of a particle

${\bf v}_{ave}=\frac{{\bf r_2}-{\bf r_1}}{t_2-t_1}$

${\bf r}_1={\bf r}(t_1)=(2.00\times 2.20+1.40){\bf\hat i}+(0.129\times (2.20)^2+1.12){\bf\hat j}$

$=5.80{\bf\hat i}+1.74{\bf\hat j}$

${\bf r}_2={\bf r}(t_2)=(2.00\times 4.05+1.40){\bf\hat i}+(0.129\times (4.05)^2+1.12){\bf\hat j}$

$=9.50{\bf\hat i}+3.24{\bf\hat j}$

So

${\bf v}_{ave}=\frac{(9.50{\bf\hat i}+3.24{\bf\hat j})-(5.80{\bf\hat i}+1.74{\bf\hat j})}{4.05-2.20}=2{\bf\hat i}+0.811{\bf\hat j}$

(b) The velocity of a particle

${\bf v}(t)={\bf r}'(t)=[(2.00t+1.40){\bf\hat i}+(0.129t^2+1.12){\bf\hat j}]'$

$=2.00{\bf\hat i}+0.258t{\bf\hat j}$

${\bf v}(2.20)=2.00{\bf\hat i}+0.258\times 2.20{\bf\hat j}=2.00{\bf\hat i}+0.568{\bf\hat j}$

The speed of a particle

$|{\bf v}(2.20)|=\sqrt{2.00^2+0.568^2}=2.08\:\rm m/s$