Distance traveled with acceleration & deceleration

  • #1
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Homework Statement


A man bungee jumps off a cliff and free falls for 3 seconds. The bungee cord stops the jumper in 5 seconds. How far did the man fall in total?


Homework Equations


X = V0t + 1/2at2

The Attempt at a Solution


Xf = X1 + X2

X1 = V1it1 + 1/2a1t12
X1 = 0 * 3 + 1/2 * -9.8 * 32
X1 = -44.1m or 44.1m (not sure if I should keep this absolute value or not...)

X2 = V2it2 + 1/2a2t22
v2i = v1f = a1 * t1
a2 = Δv/t = (v2f - v2i)/t = (v2f - (a1*t1))/t2
a2 = (0 - (-9.8*3))/5 = 5.88
X2 = (-9.8 * 3) * 5 + 1/2 * 5.88 * 25
X2 = either -73.5m, 73.5m, or 220.5m [(9.8*3)*5 + 1/2*5.88*25] ? depending on whether everything is in absolute values or not, once again, I'm not sure.

Xf = -44.1m + -73.5m
Xf = -117.6m or 117.6m

Thanks!
 

Answers and Replies

  • #2
501
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X1 = V1it1 + 1/2a1t12
X1 = 0 * 3 + 1/2 * -9.8 * 32
Is acceleration due to gravity +ve or -ve ?

Your method is basically correct . What you need to do is understand your sign convention .

Hope this helps .
 
  • #3
PeroK
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Homework Statement


A man bungee jumps off a cliff and free falls for 3 seconds. The bungee cord stops the jumper in 5 seconds. How far did the man fall in total?
If you want a quick way to solve problems like this, think about the average speed for each part of the fall. I assume the bungee deceleration is taken to be constant?
 
  • #4
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I'm glad to hear I'm not too far off on the methodology. I'm going to stick with -ve for gravity.

Yes, I believe that the deceleration is intended to be constant.
 
  • #5
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I'm glad to hear I'm not too far off on the methodology. I'm going to stick with -ve for gravity.

Yes, I believe that the deceleration is intended to be constant.
Try to find the quick way! First, what's the average speed for the free fall?
 
  • #6
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Try to find the quick way! First, what's the average speed for the free fall?
Hmmm... The only formula I know for average velocity is: Avg Velocity = Distance/Time. Distance is unknown and Avg Velocity is unknown. Is there another equation for avg velocity that I'm not aware of?
 
  • #7
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Hmmm... The only formula I know for average velocity is: Avg Velocity = Distance/Time. Distance is unknown and Avg Velocity is unknown. Is there another equation for avg velocity that I'm not aware of?
For constant acceleration from rest, average velocity is ##\frac{at}{2}##
 
  • #8
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For constant acceleration from rest, average velocity is ##\frac{at}{2}##
Interesting. Let me give it a go.

X = V1avg * t1 + V2avg * t2
X = 0.5 * a1t1 * t1 + 0.5 * a2t2 * t2
X = (0.5 * -9.8 * 3) * 3+ (0.5 * 5.88 * 5) * 5
X = -44.1 + 73.5

similar numbers as the first way, but a different sign... Certainly something else to consider! Thanks!
 
  • #9
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Interesting. Let me give it a go.

X = V1avg * t1 + V2avg * t2
X = 0.5 * a1t1 * t1 + 0.5 * a2t2 * t2
X = (0.5 * -9.8 * 3) * 3+ (0.5 * 5.88 * 5) * 5
X = -44.1 + 73.5

similar numbers as the first way, but a different sign... Certainly something else to consider! Thanks!
You need to be careful with your signs. IMHO, you are trying to plug numbers into equations without thinking enough. If gravity is negative, then the downward direction is negative, so all your displacements and velocities should be negative.

In this case, it would have been better, therefore, to take the downward direction as positive.

Here's another question: what can you say about the average velocity for the free fall and the deceleration?
 
  • #10
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I'm glad to hear I'm not too far off on the methodology. I'm going to stick with -ve for gravity.

Yes, I believe that the deceleration is intended to be constant.
You seem to be confused with your acceleration .
If this is the case , first assign your convention .

For example , let upward direction be -ve and downward +ve . Now , solve everything following this convention - Remember , all your variables ( except t ) are vectors , so pay careful attention to the sign .
 

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