Anyone show me how to work this Varying Acceleration question?

Question

A truck, part of a rocket test facility, is moving along a straight track at a steady 30 m s-1 when it is subjected to an acceleration that rises linearly from 0 to 100 m s-2 over the next 5 m of its travel. How fast is it going at this point?

thats 30 m/s speed

and 0 TO 100m/s^2 acceleration

Answer 1

The basic equation is:

v(t) = v(0) + at

v(0) = 30 m/s

Since a(t) is linear, it can be written as: a(t) = mt. (The y-intercept is 0, since there's no initial acceleration.) Therefore v(t) = 30 + at = 30 + (mt)t = 30 + mt²

We also know that a(t₁) = mt₁ = 100 m /s², where t₁ is the time it takes to travel 5 m.

But to find m and t₁, we'll have to find the distance traveled using integration.

t₁
∫ 30 + mx² dx = 30t₁ + m/3 t₁³ = 5
0

Now we've got two (UGLY!) equations in 2 unknowns:

mt₁ = 100, so m = 100/t₁

30t₁ + m/3 t₁³ = 5, so

30t₁ + 100/3 t₁² = 5

100t₁² + 90t₁ - 15 = 0

20t₁² + 18t₁ - 3 = 0

t₁ = (-18 ± √(18² - 4(20)(-3)))/(2(20))

t₁ = (-18 ± √564)/40

t₁ = (-18 ± 2√141)/40 = (-9 ± √141)/20

We're looking for positive t₁, (-9 + √141)/20 ≈ 0.1437171 s

Therefore m = 100/t₁ ≈ 695.8114.

So then

v(t₁) = v(0.1437171) = 30 + 695.8114(0.1437171)² ≈ 44.37 m/s

That was WAY harder than I thought.