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## Homework Statement

A point Q in rectilinear motion passes through the origin at t=0, and from then until 5 seconds have passed, the acceleration of Q is 6 ft/s^2 to the right. Beginning at t=5 seconds, the acceleration of Q is 12t ft/s^2 to the left. If after 2 more seconds point Q is 13 feet to the right of the origin, what was the velocity of Q at t=0?

## Homework Equations

## The Attempt at a Solution

Assume right is positive and left is negative.

t=0 to t=5

x''=6

x'=6t+(V0)

x=3t^2+(V0)t

From t=5 onwards

x''=-12t

x'=-6t^2+(V1)

x=-2t^3+(V1)t+[75+5(V0)t] (parts in brackets [] obtained by using the x=3t^2+(V0)t equation setting t=5)

we are given 13=-2t^3+(V1)t+[75+5(V0)t] at t=7s (from beginning) or t=2s (considering the equation changes at t=5)

thus plugging in t=2 (since x''=-12t begins at t=5) get:

-46=2(V1)+10(V0)

Now need another equation to solve variables. Equating the equations x'=6t+(V0) (t=0 to t=5) and the equation x'=-6t^2+(V1) (t=5 onwards)

Thus plug in t=5 for x'=6t+(V0) and t=0 for x'=-6t^2+(V1) and then equate them to get

30+(V0)=V1

Solve and get V0 = -8.83 ft/s

Book says answer is 2ft/s to the right.

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