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Velocity, acceleration and distance

  • Thread starter cash.money
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  • #1
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Homework Statement


An object is pushed from rest across a sheet of ice, accelerating at 5.0 m/s2 over a distance of 80 cm. The object then slides at a constant speed for 4.0 s until it reaches the rough section that causes the object to stop in 2.5 s.

a) What is the speed of the object when it reaches the rough section?
b) At what rate does it slow down once it reaches the rough section?
c) What is the total distance that the object slides?



Any help will be appreciated.




Homework Equations






The Attempt at a Solution



I think:
a) find v
b) find a
c) find d(distance)

I have some answers but I don't think its correct because of the rough section part of the question.
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
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OK. What do you get for the speed when it reaches the rough area?
 
  • #3
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OK. What do you get for the speed when it reaches the rough area?
I got

a) 2.8 m/s
b) -1.12 m's2
c) 15.5 m
 
  • #4
LowlyPion
Homework Helper
3,090
4
I got

a) 2.8 m/s
b) -1.12 m's2
c) 15.5 m
a) looks correct. So does b) I get a slightly different rounding.
 
  • #5
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a) looks correct. But how did you calculate b)?
b) a = (Vf - Vi)/t
a = (0 - 2.8 m/s)/2.5 s
a = - 1.12 m/s^2 answer
 
  • #6
LowlyPion
Homework Helper
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b) a = (Vf - Vi)/t
a = (0 - 2.8 m/s)/2.5 s
a = - 1.12 m/s^2 answer
It's not a big difference, but I used the √8 for v.

Now c) then is 4 * a) plus the calculation of the distance derived from slowing down over the rough. So what is the deceleration distance?
 
  • #7
14
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It's not a big difference, but I used the √8 for v.

Now c) then is 4 * a) plus the calculation of the distance derived from slowing down over the rough. So what is the deceleration distance?
This is what I did to get c)

total distance = distance when it was being pushed + distance when
it slides at constant speed for 4.0 s + distance during which it
decelerates

Dt = D1 + D2 + D3
Dt = [(2.8 m/s)^2 - 0^2]/[(2)(5.0 m/s^2)] + (2.8 m/s)(4.0 s)
+ [0^2 - (2.8 m/s)^2]/[(2)(- 1.12 m/s^2)]
Dt = 15.5 m answer
 
  • #8
LowlyPion
Homework Helper
3,090
4
This is what I did to get c)

total distance = distance when it was being pushed + distance when
it slides at constant speed for 4.0 s + distance during which it
decelerates

Dt = D1 + D2 + D3
Dt = [(2.8 m/s)^2 - 0^2]/[(2)(5.0 m/s^2)] + (2.8 m/s)(4.0 s)
+ [0^2 - (2.8 m/s)^2]/[(2)(- 1.12 m/s^2)]
Dt = 15.5 m answer
The question asks for how far it slides. I wouldn't include the .8m it was pushed.

That yields √8 * 4 = 11.314
8/(2*1.13) = 3.54
Total 14.85 m.
 

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