Question

Antiderivative problem:

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?

Answer 1

i actually changed the numbers so i can do the real problem again when i know how to do it.

I tried finding the acceleration and then antiderivative it to get the velocity..but it didnt work

Answer 2

I'm always puzzled by the conclusion "It didn't work." Please show your work and someone will help you sort it out. As a good general rule, mathematics doesn't just break. Avoiding errors is the most likely remedy.

Answer 3

F= ma
10000/ t^2 x 1/ 1000 = a
a = 10/t^2
v= 10 x t^-1 / -1
= -10 t ^-1

Answer 4

I don't see the "500 kg" in there, anywhere. All you have is Force. You need Acceleration.

Answer 5

oh .. cuz my original question is 1000 , then instead of 1/1000 , it will be 1/500

Answer 6

10000/ t^2 x 1/ 500 = a
a = 20/t^2
v= 20 x t^-1 / -1
= -20 t ^-1

Answer 7

do u know?

Answer 8

You have it. v(t) = -20/t + (Initial Velocity) for t > 1

Answer 9

but my option doesnt have that answer

Answer 10

i will screenshot u the question

Answer 11

i got v as -2m/s here. so original is 50, then it should be 48

Answer 12

We do have a little problem with this question. The acceleration doesn't act until 1 second has elapsed. Thus, v(1) should be the velocity at engine startup, 40 m/s.

Answer 13

so are the options in my m/c wrong?

Answer 14

See the other posting, please. I wasn't getting it, here.