# Force vs Time Graph

1. Mar 1, 2015

### Shoebox

1. The problem statement, all variables and given/known data
5.46 kg object moves along the x-axis. It is starting from rest. What is the velocity at 6 seconds?
2. Relevant equations
F=ma
a=f/m

3. The attempt at a solution
x=5.46(v/6)

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2. Mar 1, 2015

### BvU

Hello again, dear Shoebox,

How can I help you ? Well, first of all: by completing the problem statement with "if the force applied is as shown in the graph"

Then with an innocent: the x in part 3 doesn't appear in parts 1 and 2. So what is this mysterious symbol representing ?

It must surely be something with the dimension of kg m/s, but ?

Perhaps you want to be a bit more verbose and tell us what your game plan is ?

3. Mar 1, 2015

### Shoebox

i was trying to find the velocity by multiplying the mass by force over time. looks like i got my variables mixed up. but i realize that this is not correct.
how about using F/m=a... the force at 6 seconds is 0. the mass is 5.46.. but 0/5.46 gives me 0 acceleration

4. Mar 1, 2015

### Ashu2912

You are given the value of the force F from t=0 to t=6. So, you can find the acceleration for t=0 to t=3, then t=3 to t=5, then t=5 to t=6. Using the equations of motion in the x-direction, you can find the velocity at the end of t=3 (which will be the velocity at the beginning of t=3 to t=5), t=5 and then finally t=6.

5. Mar 2, 2015

### Shoebox

so i could use the equation for the x direction: a=f/m. for 3-5 seconds that gave me an acceleration of -0.366 m/s2. now im stuck

6. Mar 2, 2015

### BvU

What does it mean that an object has an acceleration of a m/s2 ? Your relevant equation related F and a. That's good. What is the relationship between a and the speed, v ?

---

Start with the beginning at t=0. Force = ... $\quad$ acceleration = ... $\quad$initial speed= ... $\quad$speed after 3 s is ...$\quad$So speed at t=3 s is ....

Then do the stretch from 3-5 s. Force = -2 N $\quad$ acceleration = - 0.367 m/s2$\quad$initial speed= ... $\quad$speed after 2 s is ...$\quad$So speed at t=5 s is ....

And then on to 6 s.

Last edited: Mar 4, 2015
7. Mar 2, 2015

### SammyS

Staff Emeritus
Are you familiar with the Impulse - Momentum Theorem ?

8. Mar 3, 2015

### Shoebox

i am not.. but using a kinematics chart works.