Aparticle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position is x0 = 1.1 m at t0 = 0 s . part a at 2.0 s , what is the particle's position?

The particle's position along the x-axis when t = 2.0 s is 6.4 m. It was calculated knowing that the velocity is equal to and x₀ = 1.1 m when t₀ = 0 s.  

To find the particle's position we need to remember that:

 (1)    

Where:

x: is the position  

t: is the time

The distance is then:

We know that x₀ = 1.1 m when t₀ = 0s and since need to find the position when t = 2.0s, we have:

Entering into the above equation:

Therefore, the particle's position when t = 2.0 s is 6.4 m.

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The position of the object at time t =2.0 s is 6.4 m.

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

Write an equation for x.

Substitute the equation for vₓ =2t² in the integral.

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

Substitute 2.0s for t.

The position of the particle at t =2.0 s is 6.4m

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s