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

Question #124480
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
Expert's answer

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

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