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positive infinity
May9-06, 07:20 PM
I was wondering if someone could check my work and see if its correct.

An object of mass 100.00kg moves in a straight line under the influence of a force given by F(t)=(120 N/s)t+4000N
At t=0 it is moving at 0 m/s. Determine its speed at t=5.00 s

What i did is this. I plugged 5s into the function and got 4600N. Then i used F▲V=M▲V arranged to V=(F▲T)/M
V=(4600N X 5.00s)/100.0kg
V=230. M/S

Thank you

Pyrrhus
May9-06, 09:24 PM
\int_{0}^{5} (120t + 4000)dt = m \int_{0}^{v} dv

lightgrav
May9-06, 10:53 PM
I hope that you meant F ▲t = m ▲v ... which DOES lead to ▲V = (F▲t)/m .
. . . ("change" symbols in an equation should ALWAYS balance) . . .

But why would you use the strongest Force in the 5-second duration?
Why not use the weakest Force, instead? . . . why not use the AVERAGE ?
. . . (ALWAYS multiply one's full range by the other's average) . . .