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3 years ago

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A particle initially located at the origin has an acceleration of = 1.00ĵ m/s2 and an initial velocity of i = 6.00î m/s. (a) Fin

d the vector position of the particle at any time t (where t is measured in seconds). ( t î + t2 ĵ) m (b) Find the velocity of the particle at any time t. ( î + t ĵ) m/s (c) Find the coordinates of the particle at t = 4.00 s. x = m y = m (d) Find the speed of the particle at t = 4.00 s.

1 answer:

Ira Lisetskai [31]3 years ago

7 0

Answer:

a) d = (6.00 t i ^ + 0.500 t²) m , b) v = (6.00 i ^ + 1.00 t j ^) m / s

c) d = (24.00 i ^ + 8.00 j^ ) m , d) v = (6.00 i ^ + 5 j^ ) m/s

Explanation:

This exercise is about kinematics in two dimensions

a) find the position of the particle on each axis

X axis

Since there is no acceleration on this axis, we can use the relation of uniform motion

v = x / t

x = v t

we substitute

x = 6.00 t

Y Axis

on this axis there is an acceleration and there is no initial speed

y = v₀ t + ½ a t²

y = ½ at t²

we substitute

y = ½ 1.00 t²

y = 0.500 t²

in vector position is

d = x i ^ + y j ^

d = (6.00 t i ^ + 0.500 t²) m

b) x axis

as there is no relate speed is concatenating

vₓ = v₀

vₓ = 6.00 m / s

y Axis

there is an acceleration and the initial speed is zero

$v_{y}$ = v₀ + a t

v_{y} = a t

v_{y} = 1.00 t

the velocity vector is

v = vₓ i ^ + v_{y} j ^

v = (6.00 i ^ + 1.00 t j ^) m / s

c) the coordinates for t = 4 s

d = (6.00 4 i ^ + 0.50 4 2 j⁾

d = (24.00 i ^ + 8.00 j^ ) m

x = 24.0 m

y = 8.00 m

d) the velocity of for t = 4 s

v = (6 i ^ + 1 5 j ^)

v = (6.00 i ^ + 5 j^ ) m/s

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