Answer to Question #156411 in Physics for Duc

Question

A particle has a velocity of v=(-2.0i +3.5j)m/s. The particle starts at r= ( 1.5i -3.1j)m at t=0. Give the position and acceleration as a function of time. What is the shape of the resulting path?

Answer 1

(1)

$a=\frac{dv}{dt}=\frac{d(-2i+3.5j)}{dt}=0(m/s^2)$ . Answer

(2)

$v=\frac{dr}{dt}\to dr=vdt\to r(t)=\int vdt=\int{(-2i+3.5j)dt}=$

$=-2ti+3.5tj+r_0=-2ti+3.5tj+1.5i-3.1j=$

$=(-2t+1.5)i+(3.5t-3.1)j$ . Answer

$x=-2t+1.5$ and $y=3.5t-3.1$

We get

$y=-1.75x-0.475$ - straight line. Answer