Question

Question deleted.

Answer 1

Break it into parts
First find out for the F(t)=40-20t
When this force is applied to a object at rest, how much time does it take to get to 20 m/s velocity

Answer 2

Then we have 20m/s * 4 secs
And then fine final part is
20m/s to rest applying a 60N, distance for that can be derived again from the parabollic equation
distance=1/2 a t^2+ Vot, a=acceration = F/m, Vo = initial velocity

Answer 3

acceleration*

Answer 4

\[F(t)=40-20t\]
\[a(t)=\frac{ F(t) }{ m } = 20-10t\]
\[v(t)=20t-5t^2\]
\[x(t)=10t^2-(5/3)t^3\]
you get the last two equations by integrating a(t) and v(t)

Answer 5

but, how do i use the other given figures to find the distance traveled?

Answer 6

so you need to find the time at which the force stops, and you find that using v(t):
\[20=20t-5t^2\]

Answer 7

I factorise?

Answer 8

i get t=4 ....

Answer 9

yup you solve this equation and you'll get t = 2, so the force pushes for 2 seconds and then stops for 4 seconds and then the force comes back with magnitude -60N, because it's opposing the motion (the objects decelerates to a halt).

Answer 10

yes i meant 2.. sorry..

Answer 11

are you with me so far?

Answer 12

yes.

Answer 13

great! so now we have three different equations for distance moved, the first one above, and a generic one x(t)=20t for the motion with constant velocity, and one more for last motion with the negative force:
\[F(t)=-60\]
\[a(t)=-30\]
\[v(t)=20-30t\]
\[x(t)=20t-15t^2\]

Answer 14

now we need to find the time that it takes the object to go from 20 to 0 m/s, and this is 2/3.

Answer 15

now all we need is to put 2 in the first x(t), and 4 in the second one, and 2/3 in the third x(t), and just add all the values.

Answer 16

how did you find 2/3 again?

Answer 17

by solving that last v(t) when it equals 0, so 0=20-30t, t=2/3

Answer 18

okay.

Answer 19

in the end i got 113.33 m, can you possibly check this answer?

Answer 20

yes. That is the correct answer.. thank you very much!

Answer 21

oh good, you're welcome!!