Determining initial velocity and instant velocity

[Norfonz]

Homework Statement

The acceleration of a particle is given by a_x(t) = 2.90t - 2.06

Q. Find the initial velocity v_0x such that the particle will have the same x-coordinate at time t = 4.06 as it had at t = 0.

Homework Equations

v_x = v_0x + $$\int$$a_xdt evaluated from 0 to t.
x = x₀ + $$\int$$v_xdt evaluated from 0 to t.

This is defined for straight-line motion with varying acceleration, which appears to fit my scenario.

The Attempt at a Solution

v_x = v_0x + $$\int$$(2.90t - 2.06)dt evaluated from 0 to t.
v_x = v_0x + 1.45t² - 1.03t

x = x₀ + $$\int$$(v_0x + 1.45t² - 1.03t)dt evaluated from 0 to t.
x = x₀ + $$\int$$(1.45t² - 1.03t)dt + v_0x$$\int$$dt

x = x₀ + v_0xt + (1.45t³/3) - (1.03t²/2)

Unfortunately this is as far as I got. I don't understand how to incorporate the values I was given for t. Could someone point me in the right direction please?

 

[rock.freak667]

x = x₀ + v_0xt + (1.45t³/3) - (1.03t²/2)

Unfortunately this is as far as I got. I don't understand how to incorporate the values I was given for t. Could someone point me in the right direction please?

Well from where, when t=4.06, what is x equal to ? (in terms of v_0x and x₀)

And when t=0, what is x equal to ? (in terms of v_0x and x₀)


Then just put those equal to each other and solve for v_0x

 

[kerimek]

To find the initial velocity v0x, we need to set the x-coordinate at t = 0 equal to the x-coordinate at t = 4.06. This means that x0 + v0x(0) + (1.45(0)3/3) - (1.03(0)2/2) = x0 + v0x(4.06) + (1.45(4.06)3/3) - (1.03(4.06)2/2). This simplifies to v0x = 2.90(4.06) - 2.06(4.06) = 8.65 m/s. Therefore, the initial velocity is 8.65 m/s.