A 500kg torpedo is launched from a submarine travelling at 40 m/sec. After triggering, it takes one second for the torpedo to leave the ring tube and the engine to start. After that
the force generated by the engine, in Newtons, is given by: F(x) = 10000/ t^2 , t or equal to 1 where t is measured in seconds. Recall Newton's Law that force is equal to mass times acceleration.
What is the speed of the torpedo at 6 seconds?

Question

A 500kg torpedo is launched from a submarine travelling at 40 m/sec. After triggering, it takes one second for the torpedo to leave the ring tube and the engine to start. After that
the force generated by the engine, in Newtons, is given by: F(x) = 10000/ t^2 , t> or equal to 1 where t is measured in seconds. Recall Newton's Law that force is equal to mass times acceleration.
What is the speed of the torpedo at 6 seconds?

tkhunny · Answer

It's really annoying when you miss stuff that easy.  $$F(t) = \frac{10000}{t^2}$$  This gives $$a(t) = \frac{10000}{500\cdot t^{2}} = \frac{20}{t^{2}}$$  Rather than worry about the velocity at all times, why not just answer the question?  $$\int_{1}^{6}a(t)\;dt + 40 $$

anonymous · Answer

oh..ok.. i get it this way. .but i thought just finding velocity will work out too..umm ok thnks!

tkhunny · Answer

We're not used to it starting at t = 1.  This was the point of most confusion.  $$v(t) = v_{0} + \int_{1}^{t}\frac{20}{t^{2}}\;dt = ??$$  It's that last piece, that is usually zero, that was confusing us both.

anonymous · Answer

omg i got it!

anonymous · Answer

what i did is using F(b) - F(a)

anonymous · Answer

and i did got the answer, haha thanks!

tkhunny · Answer

Sorry, that's bad form.  I should have switched variables in that inegral.  The t^2 and the dt don't quite mean the same thing as the other two 't's.