Question

The position vs time of a object is given by

$LaTeX: \small \vec{r} = (3 m - 5 \frac{m}{s} t)~ \hat{i} +(2 m + 3 \frac{m}{s^2} t^2 )~\hat{j} - 4 \frac{m}{s^2} t^2 ~\hat{k}$

1.What is the y velocity at 4.5 seconds?

2.At what time is the speed equal to 15 m/s?

The velocity of an object vs. time is given by

$LaTeX: \vec{v} = (3 \frac{m}{s} - 2 \frac{m}{s^2 } t) \hat{i} + 1.5 \frac{m}{s^3} t^2~\hat{j}$

3. What is the acceleration at t = 2s?

Answer 1

r vector = (3m - 5 m/s t)i^^+ (2m + 3 m/s^2 t^2)j^^- 4 m/s^2 t^2 k^^v vector = dr vector/dt = (0 - 5 m/s)i^^+ (0 + 3 times 2 m/s^2 t)j^^- 4 times 2 m/s^2 t k^^= (-5 m/s)i^^+ (6 m/s^2 t)j^^- 8 m/s^2 tk^^v vector = -5 m/s i^^+ 6 m/s^2 t j^^- 8 m/s^2 t k^^(1) \r\n From equation (1) at t = 4.5 sec V_y = 6 m/s^2 t = 6 m/s^2 times 4.5 sec = 27 m/sec. From equation (1) Speed V = squareroot (-5)^2 + (6t)^2 + (8t)^2 = 15 V = squareroot 25 + 100t^2 = 15 Squaring both side 25 + 100t^2 = (15)^2 100t^2 = (15)^2 - 25 = 200 t^2 = 2 t = 1.414 sec at t = 1.414 sec speed = 15 m/s v vector = (3 m/s - 2 m/s^2 t)i^^