# Stopping Distance Question

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1. Apr 6, 2016

### Johny Prime

1. The problem statement, all variables and given/known data
Having trouble with the concept here and have gotten stuck!

A man is pushing a mobile machine down a corridor at a steady speed of 1.2 m/s. The machine’s mass is 55 kg. Another person who has been walking in front of the machine has suddenly stopped. If the man is able to apply sufficient force to stop the machine in 1.0 s., will the man avoid a collision if the person is 1.0 m ahead of the machine? (Assume that the man is applying the same magnitude of force throughout that time.)

Velocity (initial) = 1.2 m/s
Mass = 55 kg
Velocity is at 0 after 1 second
Distance between is 1 meter

2. Relevant equations
• Acceleration = (Vf-Vi)/time
• Force = acceleration X mass
• Velocity = Distance/time
3. The attempt at a solution
Vf= 0 (stopped)
Vi= 1.2 m/s
So now finding the acceleration (deceleration): (0-1.2m/s)/1s = -1.2 m/s2
Can now find the force required to stop: Force = -1.2 X 55 kg
Force = -66 newtons

This is where I am stuck. I am not sure if I am supposed to use : Force X Distance = 1/2 mv2 formula (kinetic energy and work etc) or I am missing something obvious? If my work is correct so far what should be my next step?

2. Apr 6, 2016

### Borg

It's probably a lot easier to step back and think about it differently. If he stops in one second, what do you think the average speed would be during that time?

3. Apr 6, 2016

### Let'sthink

I think the problem must mention the coefficient of kinetic friction. Assume that it is given and one can find the limiting force, say of magnitude, f which man is applying to move the machine at a constant speed of whatever value. Now man wants to stop the machine and s applies the same force now in opposite direction helping friction. So the force applied on the machine will be 2f = 66 as correctly calculated by you. So f > 33 N.

4. Apr 6, 2016

### Johny Prime

The average velocity would be -1.2 m/s, wouldn't it? I'm still not seeing the connection between this and finding the distance.

5. Apr 6, 2016

### Staff: Mentor

You're given that the machine stops in 1.0 seconds going from 1.2 m/s to 0 m/s in that time. If the man applies a constant force to achieve this then the acceleration will be constant and all the SUVAT formulas are available to use. Or just draw a velocity vs time graph and reminisce about that lesson where the prof talked about the area under a curve...

6. Apr 6, 2016

### Johny Prime

Wait, I think i've figured it out. The average velocity would be 0.6 m/s right? Meaning that he travels 0.6 m? Is that correct?

7. Apr 6, 2016

Yes.

8. Apr 6, 2016

### Johny Prime

Thank you so much for your help, I think I understand it now.

9. Apr 6, 2016

### Staff: Mentor

You're welcome.

10. Apr 7, 2016

### Let'sthink

Did down mean down the slope. Then it is ok I can understand, otherwise just suvat based questions without description of the situation is not good for developing correct concepts.

11. Apr 7, 2016

### rodgerr

I dont understand how you worked out the average velocity to be 0.6 m/s. Could you please explain this?

12. Apr 7, 2016

### CWatters

Hint:
Draw a graph of velocity vs time.
At t=0 v=1.2m/s
At t=1 v=0m/s
At constant acceleration the slope is constant/straight line between these two points.

13. Apr 7, 2016

### CWatters

You can also solve this problem using the SUVAT equation to calculate the stopping distance..

S = (U+V)t/2
= (1.2+0)*0.5
= 0.6m

The (U+V)/2 part is the average velocity.