Question

Suppose that the position vector for a particle is given as a function of time by vector r (t) = x(t)î + y(t)ĵ, with x(t) = at + b and y(t) = ct2 + d, where a = 1.30 m/s, b = 1.45 m, c = 0.129 m/s2, and d = 1.16 m. (a) calculate the average velocity during the time interval from t = 2.05 s to t = 4.25 s.

Answer 1

position in x direction is given as

$x = at + b$

a = 1.30

b = 1.45

$x = 1.30 + 1.45 t$

$x_1 = 1.30 + 1.45*2.05 = 4.2725m$

$x_2 = 1.30 + 1.45*4.25 = 7.4625 m$

now average velocity in x direction is given as

$v_x = \frac{x_2 - x_1}{t}$

$v_x = \frac{7.4625 - 4.2725}{4.25 - 2.05}$

$v_x = 1.45 m/s$

Now similarly in Y direction position is given as

$y = ct^2 + d$

here

c = 0.129

d = 1.16

$y = 0.129t^2 + 1.16$

$y_1 = 0.129*2.05^2 + 1.16 = 1.702$

$y_2 = 0.129*4.25^2 + 1.16 = 3.490$

now by the average speed formula

$v = \frac{y_2 - y_1}{t}$

$v = \frac{3.49 - 1.702}{4.25 - 2.05}$

$v_y = 0.81 m/s$