Deceleration with weight and a dent? Confused

In summary, the conversation is about finding the magnitude of deceleration of a 27 pound meteorite that struck a car with a speed of 520 m/s, leaving a dent 27 cm deep in the trunk. The conversation includes discussions about using equations such as KE = F * d and Vf^2=Vi^2+(2ad) to calculate the deceleration, as well as converting pounds to Newtons and Newtons to kilograms. The conversation also covers the concepts of displacement and direction in relation to the acceleration.
  • #1
7Kings
9
0
Ive been trying to fiure out how to get this answer but am just have some serious trouble.

A 27 pound meteorite struck a car, leaving a dent 27 cm deep in the trunk. If the meteorite struck the car with a speed of 520 m/s, what was the magnitude of its deceleration, assuming it to be constant?

I just can't figure out how this is done. To find the magnitude of deceleration i believe i would need to divide the 520 m/s by some other time in seconds yet...there is no other time in seconds. Also, I haven't an idea what I am supposed to do with the pounds and dent. Is there a way to find out seconds from those or something? I am really stuck here. If someone could just guide me along or give me a formula to figure it how to do this it would be greatly appreciated, as I just cannot find one!
 
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  • #2
What equations do you know?
 
  • #3
You can calculate the Kinetic Energy of the meteorite prior to striking the car. Use KE = F * d to get the average force that the dent acted on the
meteorite during the impact. Once you have the average force you should
be able to calculate the deceleration.

A 27 lb object traveling at nearly the speed of a rifle bullet and a dent??
You've got to be kidding!
 
  • #4
Don't forget to convert pounds to Newtons, and Newtons into kilograms.
 
  • #5
What does F, and d stand for? I know KE is the kinetic energy. I did the conversion of pounds to Kilograms. It would be 12.246994 kg.

And the only equations i learned where the four kinetics equations, but i don't see how they could possibly help me yet as none of them have anything to do with mass.
 
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  • #6
F is the force in Newtons in the metric system
d is the distance in meters

F * d is the work in Joules done in stopping the meteorite which is
also equal to the Kinetic Energy of the meteorite at impact
Also, there is no use in using more significant figures than 12.2 Kg
for the mass since you are limited to 2 significant figures in the depth
of the dent.
 
  • #7
7Kings said:
And the only equations i learned where the four kinetics equations, but i don't see how they could possibly help me yet as none of them have anything to do with mass.
You don't need the mass. You are given all the information needed to solve this as a kinematics problem.

Hints:
What distance does the object travel (during its deceleration)?
What are its initial and final speeds?
 
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  • #8
Thats what I am missing. I don't see where it tells me the distance the object travels, or how i can figure that out. Also, I don't see its inital speed. I just know its final speed is 520 m/s
 
  • #9
Well:
1. Is the object decelerating before it hits the car?
2. How far does the object travel while it's in contact with the car?
 
  • #10
Actually, I've just made some progress. I've found out i need to use the equation

Vf^2=Vi^2+(2ad)

but I am not sure what i do next.
Was i wrong before and is the INITIAL speed 520 m/s, the FINAL speed 0 m/s?

That would have 0^2=520^2+2ad but...i don't know what the displacement is.

Oh wait, Displacement=final v-inital v. so now i have

0^2=520^2+2a(0-520)
0=270400+2a(-520)
0=270400-1040a
1040a=270400
a=260 m/s/s

Am i right or did i place something wrong? Shouldnt the acceleration be negative? Since its deceleration I am getting?

Sorry about all the editing to this but i think i got it now. I just solved for A and got:
A=(Vf^2-Vi^2)/2d
A=(0-520^2)/2*520
A=-260 m/s/s
 
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  • #11
Ok. I don't want to keep on posting my random ideas but this time I am almost positive i did it right.
A=(Vf^2-Vi^2)/2d
A=(0-550^2)/(2*-.27) (Is the -.27 negative becuase its a dent so it went like..inwards kinda?)
Then you have A=500740.7407 m/sec^2 as your final answer?
 
  • #12
Sorry, sorry guys. I've just got one more thing, and i finished that question. Just the next one, its basically the same thing but I am not sure about one part of it:

Coasting due south on your bicycle at 8.0 m/s, you encounter a sandy patch of road 6.0 m across. When you leave the sandy patch your speed has been reduced to 6.6 m/s. Assuming the bicycle slows with constant acceleration, what was its acceleration in the sandy patch? Give both magnitude and direction.

So once again we use the same formula A=(Vf^2-Vi^2)/2d
A=(6.6^2-8^2)/(0-6) A=3.406 m/sec^2 due north OR
A=(6.6^2-8^2)/(6-0) A=-3.406 m/sec^2 due north

im not sure which way the displacement part of it goes but I am pretty sure my direction part of it is right, since its decelerating the direction would be the opposite. And i know, Displacement=final position-initial position...im just not sure what 6 is!
 
  • #13
7Kings said:
Ok. I don't want to keep on posting my random ideas but this time I am almost positive i did it right.
A=(Vf^2-Vi^2)/2d
A=(0-550^2)/(2*-.27) (Is the -.27 negative becuase its a dent so it went like..inwards kinda?)
Then you have A=500740.7407 m/sec^2 as your final answer?
That's OK, but please round it off to two significant figures.

If you are worried about the sign of your answer, you must use a sign convention consistently. If you call up positive and down negative, then:
Vi = -520 m/s
d = change in position = -0.27 m - 0 m (calling the initial position at impact = 0)

Using this convention the acceleration will end up positive (upwards), which should make sense. After all, the car pushes up on the meteorite, bringing it to rest. Since the acceleration is opposite to the velocity, the term "deceleration" is used (which just means it slows down).
 
  • #14
7Kings said:
So once again we use the same formula A=(Vf^2-Vi^2)/2d
A=(6.6^2-8^2)/(0-6) A=3.406 m/sec^2 due north OR
A=(6.6^2-8^2)/(6-0) A=-3.406 m/sec^2 due north
Realize that a negative acceleration due north means the same as saying that the acceleration is due south.

im not sure which way the displacement part of it goes but I am pretty sure my direction part of it is right, since its decelerating the direction would be the opposite. And i know, Displacement=final position-initial position...im just not sure what 6 is!
Again, pick a sign convention like I explained in my last post. Call the initial position = 0.

But you really should be able to write down the direction of the acceleration immediately, without fussing with sign conventions or equations. Hint: Is the bike slowing down or speeding up? Use the equation to get the magnitude of the acceleration, but use your understanding to assign the direction.

Of course, if you are strict with the sign convention, the equation will automatically provide the correct sign.
 
  • #15
So then are you saying that 3.406 m/s^2 due north = -3.406m/s^2 due south, since its going south?
 
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  • #16
7Kings said:
So then are you saying that 3.406 m/s^2 due north = -3.406m/s^2 due south, since its going south?
No. I'm saying that taking negative two steps forward is the same as taking two steps backward. :wink:

The magnitude of a vector is always positive (or zero). So when you specify a vector, such as acceleration, by giving its magnitude and direction, give a positive number and a direction (never a negative number).
 

What is deceleration?

Deceleration is the rate at which an object's velocity decreases over time. It is the opposite of acceleration, which is the rate at which an object's velocity increases over time.

How does weight affect deceleration?

The weight of an object does not directly affect its deceleration. Instead, it is the mass of an object that determines how much it will resist changes in its velocity. An object with greater mass will require more force to decelerate compared to an object with lesser mass.

What is a dent and how does it affect deceleration?

A dent is a deformity or indentation in an object's surface. It can affect deceleration by changing the object's shape and altering its aerodynamics, which can increase drag and slow down the object's motion. It can also change the distribution of mass, which can impact the object's center of gravity and stability during deceleration.

Why am I confused about deceleration with weight and a dent?

You may be confused about deceleration with weight and a dent because these factors can interact in complex ways to affect an object's motion. Deceleration is also a counterintuitive concept, as it is often associated with slowing down or stopping, but it is actually the rate of change in velocity. Additionally, the effects of weight and a dent on deceleration may vary depending on other factors such as the object's speed and the surface it is moving on.

How can I calculate deceleration with weight and a dent?

Calculating deceleration with weight and a dent requires the use of Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration. By measuring the object's mass and the force acting on it during deceleration, you can calculate the deceleration using the formula a = F/m. However, the presence of a dent may complicate this calculation, as it may affect the object's mass and the force acting on it.

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