Question

The acceleration of a particle is constant. At t = 0, the particle is at the origin and the velocity of the particle is vo = vji+ v2j. At time t = T, the velocity of the particle is v = v3j. Here v1, v2, and v3 are constants with dimensions of length divided by time. All answers should be written in terms of vj, v2, V3, T, and the unit vectors i and j.

Part (a) What is the particle's acceleration vector?
Part (b) What is the particle's position vector at t= 2T?
Part (c) What is the particle's velocity vector at t =2T?

Answer 1

a) a =
`- (v_1)/(T)` i ^ +
`(v_3 - v_2)/(T)` j^, b) r = 2 v₃ T j ^, c) v = -v₁ i ^ + (2 v₃ - v₂) j ^

Step-by-step explanation:

This is a two-dimensional kinematics problem

a) Let's find the acceleration of the body, for this let's use a Cartesian coordinate system

X axis

initial velocity v₀ₓ = v₁ for t = 0, velocity reaches vₓ = 0 for t = T, let's use

vₓ = v₀ₓ + aₓ t

we substitute

for t = T

0 = v₁ + aₓ T

aₓ = - v₁ / T

y axis

the initial velocity is
`v_(oy)` = v₂ at t = 0 s, for time t = T s the velocity is v_{y} = v₃

v₃ = v₂ + a_{y} T

a_{y} =
`(v_3 - v_2)/(T)`

therefore the acceleration vector is

a =
`- (v_1)/(T)` i ^ +
`(v_3 - v_2)/(T)` j^

b) the position vector at t = 2T, we work on each axis

X axis

x = v₀ₓ t + ½ aₓ t²

we substitute

x = v₁ 2T + ½ (-v₁ / T) (2T)²

x = 2v₁ T - 2 v₁ T

x = 0

Y axis

y =
`v_(oy)` t + ½ a_{y} t²

y = v₂ 2T + ½
`(v_3 - v_2)/(T)` 4T²

y = 2 v₂ T + 2 (v₃ -v₂) T

y = 2 v₃ T

the position vector is

r = 2 v₃ T j ^

c) the velocity vector for t = 2T

X axis

vₓ = v₀ₓ + aₓ t

we substitute

vₓ = v₁ -
`(v_1)/(T)` 2T = v₁ - 2 v₁

vₓ = -v₁

Y axis

`v_(y)` = v_{oy} + a_{y} t

v_{y} = v₂ +
`( v_3 - v_2)/(T)` 2T

v_{y} = v₂ + 2 v₃ - 2v₂

v_{y} = 2 v₃ - v₂

the velocity vector is

v = -v₁ i ^ + (2 v₃ - v₂) j ^