EK 1001 # 56 Physics

[shree1919]

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A particle moving at 5 m/s reverses its direction in 1 s to move at 5 m/s in the opposite direction. If its acceleration is constant, what distance does it travel?

Can someone please explain this to me?

 

[dmf2682]

A particle moving at 5 m/s reverses its direction in 1 s to move at 5 m/s in the opposite direction. If its acceleration is constant, what distance does it travel?

Can someone please explain this to me?

Yeah. Draw a velocity v time curve. Start at 5 m /s. Slope is constant so you know its linear. At the 1s mark place a point at -5m/s . Line between them. Since you're looking for distance and not displacement its going to be the area under the two triangles added up. If it were displacement its 0. Should get 1 m.

 

[milski]

Δv=-5m/s - 5m/s=-10m/s
a=Δv/t=-10m/s / 1s = -10m/s^2

Direction will get reversed when v is 0, which is at t=1/2
Distance from when it started to decelerate to 0 is d=vt+a/2t^2=5*1/2-10/2*1/4=5/4. Distance in the opposite direction is the same, so total distance will be 5/4+5/4=5/2=2.5 m.

 

[chiddler]

could this question be done without breaking it into two parts as milski did via

(vi + vf)/2 *t = d

?

 

[dmf2682]

Δv=-5m/s - 5m/s=-10m/s
a=Δv/t=-10m/s / 1s = -10m/s^2

Direction will get reversed when v is 0, which is at t=1/2
Distance from when it started to decelerate to 0 is d=vt+a/2t^2=5*1/2-10/2*1/4=5/4. Distance in the opposite direction is the same, so total distance will be 5/4+5/4=5/2=2.5 m.

Lol woops yeah I dunno why I thought 1. Its 2.5

 

[MedPR]

could this question be done without breaking it into two parts as milski did via

(vi + vf)/2 *t = d

?

Yes, but only because acceleration is constant. The equation you're using is one of those extra ones that not everyone has memorized.

 

[milski]

could this question be done without breaking it into two parts as milski did via

(vi + vf)/2 *t = d

?

You have to break it at least in the parts of moving in the positive and the negative direction if you want the distance. Otherwise you'll get the displacement.

But you can do (5+0)2*1/2=5/4 from 5 to 0 and then just double it.

 

[dmf2682]

could this question be done without breaking it into two parts as milski did via

(vi + vf)/2 *t = d

?

Not like that. Average velocity is 0.

The other poster didn't need to break it into 2.

 

[MedPR]

You have to break it at least in the parts of moving in the positive and the negative direction if you want the distance. Otherwise you'll get the displacement.

But you can do (5+0)2*1/2=5/4 from 5 to 0 and then just double it.

Not like that. Average velocity is 0.

The other poster didn't need to break it into 2.

Oh, I thought we were talking about 5/2 *.5 = d and -5/2*.5 = d, then adding the displacements together.

That still breaks it into two parts though 😀 sorry.

 

[dmf2682]

Do the graph thing below. The math works out, I just wrote the wrong answer. If you're a visual learner graphs are the best thing for physics.

 

[milski]

Yes, the graph works too. If calculating the area of two triangles is splitting in two or not is just a matter of semantics. 👍

 

[chiddler]

Do the graph thing below. The math works out, I just wrote the wrong answer. If you're a visual learner graphs are the best thing for physics.

yeah but who is going to draw a graph to get an answer on the mcat 😛

 

[MedPR]

yeah but who is going to draw a graph to get an answer on the mcat 😛

It wouldn't be my first choice, but if I have extra time I will. I've been trying to think of a few different ways to solve various problems just in case .