please i need help asap Problem 1 The acceleration of a particle moving only on

Question

please i need help asap

Problem 1

The acceleration of a particle moving only on a horizontal xy plane is given by a=3ti+4tj, where a is in meters per seconds squared and t is in seconds, at t=0, the position vector r=(20.0m)i+(40.0m)j locates the particles, which then has the velocity vector v=(5.00m/s)i+(2.00m's)j. at t=4.00s, what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the x axis?

Problem 2

A particle starts from the origine at t=0 with a velocity of 8.0j m/s and moves in the xy plane with constant acceleration (4.0i+2.0j)m/s^2. when the particle's x coordinate is 29m, what are its (a) y coordinate and (b) speed?

Explanation / Answer

v(t) = vx(t)i + vy(t)j = ? a(t) dt

=> vx(t) = ? ax(t) dt = ? 3t dt = (3t^2 / 2 + C1)

since vx(0) = 5 => C1 = 5

=> vx(t) = 3t^2 / 2 + 5 <<<<<<<

do the same for vy(t) :

vy(t) = 4(t^2 / 2) + C2

C2 = vy(0) = 2

hence vy(t) = 2t^2 + 2 <<<<<<<<

so v(t) = (1.5 t^2 + 5) i + (2t^2 + 2) j <<<<<<<<<

do the same for the position function d(t) :

a) dx(t) = ? vx(t) dt = ? (1.5 t^2 + 5)dt = 1.5 (t^3/3) + 5t + d1

dx(0) = d1 = 20

hence dx(t) = 0.5 t^3 + 5t + 20 <<<<<<

b) do the same for dy(t) :

dy(t) = ? vy(t) dt = ? (2t^2 + 2) dt = (2/3)t^3 + 2t + d2

d2 = dy(0) = 40 m

so dy(t) = (2/3)t^3 + 2t + 40

hence ,

d(t) = (0.5 t^3 + 5t + 20) i + ( (2/3)t^3 + 2t + 40) j


angle = arctan (vy/vx) = arctan [(2t^2 + 2) / (1.5 t^2 + 5)]

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