A particle has a constant acceleration of vector a = (5.9 m/s2)i + (3.8 m/s2)j . At time t = 0, the velocity is zero and the position vector is vector r0 = (11.1 m)i.

(a) Find the velocity and position vectors as functions of time t.

V= m/s

r= m

(b) Find the equation of the particle's path in the xy plane. (Use the following as necessary: x.)

y= m

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cjmejiab cjmejiab · Answer 1 · 2019-11-12T05:05:58+01:00

To find the solution to this problem we need the concept relate to equations of kinematics to find the velocity and position, that is

Part a) We know the value of the Acceleration so, to find the velocity find the relation between acceleration and velocity, which is given by,

[tex]\vec{V_f} = \vec{V_i}+\vec{a}t[/tex]

[tex]\vec{V_f} = 0 +(5.9\hat{i}+3.8\hat{j})t[/tex]

Then the velocity is given by,

[tex]\vec{V_f} = 5.9t\hat{i}+3.8t\hat{j}[/tex]

Part b)To find the position we apply the same Kinematic equation but for position, who is given by,

[tex]\vec{r_f} = \vec{V_i}t+\frac{1}{2}\vec{a}t^2[/tex]

Replacing,

[tex]\vec{r_f} = 0+\frac{1}{2}(5.9\hat{i}+3.8\hat{j})t^2[/tex]

[tex]\vec{r_f} = 2.95t^2\hat{i}+1.9t^2\hat{j})[/tex]

Part c) We find the equation of the particle's path with positon equation, that is,

[tex]\vec{r_f} = 2.95t^2\hat{i}+1.9t^2\hat{j})[/tex]

Where,

[tex]y= 1.9t^2[/tex]

[tex]x= 2.95t^2[/tex]

[tex]\frac{y}{x} = \frac{1.9}{2.95}[/tex]

Then,

[tex]y= 0.644x[/tex]