Answer to Question #124480 in Mechanics | Relativity for Mohamed sayga

Question

A particle initially located at the origin has an acceleration of-2.00j m/s2 and an initial velocity of7.00i m/s (a) Find the vector position of the particle at any time t (where t is measured in seconds) t2 j) m (b) Find the velocity of the particle at any time t tj) m/s (c) Find the coordinates of the particle at t 4.00 s (d) Find the speed of the particle at t 4.00 s m/s

Answer 1

Initial position of the particle (0,0)

acceleration, $a_y = -2\hat j$ m/s²

initial velocity, $u_x = 7 \hat i$ m/s

Distance covered by particle along x-axis in time t,

$x = u_xt = 7t$

Distance covered by particle along y-axis in time t,

$y = \frac{1}{2} a_yt^2= 0.5 * (-2)t^2 = - t^2$

Position of the particle at time t,

$\vec{r} = 7t\hat{i} -t^2\hat{j}$ m

Velocity along x axis will remains the same while along y-axis it will change due to acceleration at any time,

$v_x = 7\hat{i}$ $v_y = a_yt \hat j= -2t \hat j$

so velocity vector will be,

$\vec{v} = 7\hat i- 2t\hat j$ m/s

Co-ordinates of particle at 4 second,

$\vec{r} = 7t\hat{i} -t^2\hat{j} = (7*4)\hat{i} -(4)^2\hat{j} = 28 \hat i - 16 \hat j$

x-co-ordinates is 28 while y co-ordinate is -16

Velocity at 4s

$\vec{v} = 7\hat i- 2t\hat j = 7\hat i- (2*4)\hat j = 7\hat i- 8\hat j$

speed will be $s = \sqrt{7^2 + 8^2 } = \sqrt{49+64}=10.63$ m/s approx