Classical mechanics: Force versus time graph

[REVIANNA]

Homework Statement

A particle of mass m=4.0 kg is moving along the x-axis. The particle is being acted upon by a variable single force F⃗ , directed along the x-axis. At t=0 s, the particle is moving at v_0=−3 m/s.

What is the first time t>0 when the particle comes to a stop momentarily?

The Attempt at a Solution

because I know the initial velocity ,I know the initial momentum and I know the final momentum too(as the particle stops temporarily) but the force is variable otherwise I would have equated change in momentum to the impulse imparted.and found the time t.

what should I do?

 

[gneill]

What does area under a Force vs Time curve represent?

 

[REVIANNA]

What does area under a Force vs Time curve represent?

it represents impulse.

$mv_f-mv_i=J$
$0-m(-3)=∫F. dt$

 

[cnh1995]

it represents impulse.

$mv_f-mv_i=J$
$0-m(-3)=∫F. dt$

So you'll have to calculate the value of t at which the area of the curve becomes 12.

 

[REVIANNA]

So you'll have to calculate the value of t at which the area of the curve becomes 12.

how will I do it? I don't have the force as a fn of time.

 

[cnh1995]

I don't have the force as a fn of time.

You do.

First you need to formulate F as a function of time between various intervals.

 

[REVIANNA]

you need to formulate F as a function of time between various intervals.

you mean from 0 to 1 sec its 2.5
from 1 to 4 its 12

but the answer is 3.375

 

[cnh1995]

answer is 3.375

Indeed it is. That's the final answer.

you mean from 0 to 1 sec its 2.5
from 1 to 4 its 12

No. From t=0 to t=1, what is F(t) and so on..

 

[cnh1995]

Indeed it is. That's the final answer.

No. From t=0 to t=1, what is F(t) and so on..

Ok right, if you are talking in terms of areas directly..

 

[REVIANNA]

if you are talking in terms of areas directly..

area under the f/t curve is all I know what else are you saying.also I am fairly new to calculus

 

[cnh1995]

you mean from 0 to 1 sec its 2.5
from 1 to 4 its 12

Ok. So, from 0 to 4, the area is 14.5. So, t must be between 1 to 4. Agree? Because total area is 12 and out of that, 2.5 is already in between 0 to 1.

 

[REVIANNA]

Ok. So, from 0 to 4, the area is 14.5. So, t must be between 1 to 4. Agree?

yes totally because of the areas.

 

[cnh1995]

yes totally because of the areas.

So, from t=1 to t=4, you need to find at what value of t will the area be 9.5(i.e. 12-2.5). Can you do it?

 

[cnh1995]

yes totally because of the areas.

You know its a rectangle from t=1 to t=4. You know the height. I believe the answer is in plain sight. Good luck..

 

[REVIANNA]

answer is in plain sight

$∫F.dt+F_c(t-1)=-mv_i$
(the integral is form t=0 to t=1) (c means const)
$2.5+4(t-1)=-4*(-3)$
$t=3.375$

@cnh1995
thank you