# 1-D Kinematics

#### kingwinner

Registered Senior Member
1) A balloon is ascending at 9.0m/s and has reached a height of 80m above the ground when it releases a package. How long does the package take to reach the ground?

There is only one question I'd like to ask. Would the initial velocity of the dropped package be 0m/s since it is dropped?

2) A rocket-driven sled running on a straight, level track can attain a velocity of 1600km/h in 1.8s, starting from rest. Under the last 2.00km of the track there is a trough of water. The "brakes" of the rocket consist of a scoop that displaces water from the trough, stopping the sled. Assuming that the sled stops at the end of the trough, what is the deceleration of the sled, expressed in terms of g? (g=9.8m/s^2)

I don't know how to do question 2. Can someone give me some hints? Thank you!

1) The problem says the package is "released" i.e. its just let go, so I would interpret this to mean that the package is released at rest with respect to the balloon. What then is the package's velocity relative to the Earth?

2) For uniformly accelerated motion, how is the velocity related to the acceleration? What about the position? Remember that acceleration can be negative as in the case of breaking.

1) I would say that the initial velocity of the package would be 9.0 m/s since the balloon didn't stop to release the package.

2) You know that the initial velocity is 1600 km/h, the final velocity is 0 (or equivalently the change in velocity is -1600 km/h), and the change in position is 2 km (or equivalently the initial position is 0 and the final position is 2 km). From that you can calculate the acceleration. Then all you have to do is convert from km/h^2 to g.

-Dale

I would answer the question with both 9.0 and 0 velocity, since the problem didn't state it very clearly, then ask my teacher what one he/she is looking for then put an x over the wrong one. (it can easily be done if leave the initial velocity term in the solution until the last step.)

1) I know that if the balloon is stationary, when it drops a package, the initial velocity of it would be 0m/s.
But if the balloon is rising up at 9.0m/s, and drops the package, would the initial velocity of the package still be 0m/s or would it be 9.0m/s [up]? If it is 9.0m/s[up], I don't understand...the package is released, not thrown upward or downward, why would it have velocity that is not 0?

If a person on the earth sees the releasing of the package as the balloon is rising, would he see that the package initially is also going upwards (at 9.0m/s) instead of downwards?

2) I got
v(initial)=0
v(final)=444.4m/s
delta t=1.8s

From that I can calculate the acceleration of the sled =246.667m/s^2

But for the rest of the question:
"...Under the last 2.00km of the track there is a trough of water. The "brakes" of the rocket consist of a scoop that displaces water from the trough, stopping the sled. Assuming that the sled stops at the end of the trough, what is the deceleration of the sled, expressed in terms of g?"

Only two values are given: the displacement =2.0km, and the final velocity is 0, I don't know the time interval, nor the initial velocity! With only two known values, how can I calculate the deceleration of the sled?

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On 1), think of it this way. Prior to the release of the package, the balloon is moving upwards at 9 m/s, and the package is moving upwards at 9 m/s because its attached to the balloon. When the package is released it isn't going to suddenly stop in an instant, so it will initially be moving upwards at the same 9 m/s as it was a moment before release.

On 2), you do know the initial velocity where here what is meant by initial velocity is the velocity the sled has right before it enters the water trough. What is it? You now know the initial velocity, the final velocity, and the length of the track. This is enough to figure out the stopping time and the acceleration.

1) Oh, I see now!

2) What is the initial velocity?
After the sled reached 1600km/h, is it still keep accelerating or maintaining a constant velocity of 1600km/h? If keep accelerating, for how long before it enters the water trough?

I assume that the rocket turns off after 1.8 s and the sled glides along at constant velocity until it reaches the water trough. Otherwise, you would need to know the full length of the track to figure out how fast the sled is going when it reaches the trough, but you don't have this information. This makes the speed of the sled 1600 km/h when it first enters the trough, yes?

2) Thanks for explaning! But I don't think the question did mention all this...how did you know? Does the sled enter the water trough right after it reaches 1600km/h?

3) Two trains, one travelling at 100km/h and the other at 128km/h, are headed towards one another along a straight, level track. When the trains are 1.2km apart, each engineer simultaneously sees the other's train and applies the brakes. Both trains have equal, constant decelerations of 0.9m/s^2. Will there be a collision.

[I understand how to calculate this. The only problem is should I use the rounded or unrounded answer for determining if there is a collision? If the sum of the magnitudes of the displacement of the two trains 1196m, will they collide? If rounded to the correct number of significant digits, 1196m=1x10^3m
If I do have to round, what if the sum of the magnitudes of the displacement of the two trains 1311m? If rounded to the correct number of significant digits, 1311m=1x10^3m as well, will there be a collision?]

4) A convertible with its top down drove towards the entrance of an underground garage with a velocity of 24km/h. A window cleaner on a scaffold directly above the entrance accidentally kicked a bucket of water off a moving scaffold. The scaffold at that moment was 9.0m vertically higher than the top of the car and the car was 12m from the entrance. The scaffold was moving up at 1.5m/s. Did the dirver get wet?

[In what case would the driver get wet? Is it true that he would get wet if the time of water = time of car and are EXACTLY the same? to how many significant digits?
I got:
time of water=1.52s
time of car=1.80s
Do I have to round to 2 significant digits before determining if the driver will get wet, or should I use the unrounded answer?]

For 3) the honest answer is that there is not enough precision in the measurement of deceleration to be able to determine if a collision will occur or not (for exactly the reasons you pointed out). I think they don't want you to round so much, so I think that the answer they want is no collision.

For 4) it looks like all the data has 2 significant digits so I would think that 1.8s>1.5s means that the water hits his windshield or something.

-Dale

hahahaha, kingwinner, I thought that first problem sounded familiar! you are using Halliday/Resnik/Walker Fundamentals of physics aren't you? well, maybe not, I suppose many classes use these examples.

DaleSpam said:
For 3) the honest answer is that there is not enough precision in the measurement of deceleration to be able to determine if a collision will occur or not (for exactly the reasons you pointed out). I think they don't want you to round so much, so I think that the answer they want is no collision.

For 4) it looks like all the data has 2 significant digits so I would think that 1.8s>1.5s means that the water hits his windshield or something.

-Dale
3) What if I really got that the sum of the magnitudes of the displacement of the two trains 1311m, can I just use the unrounded answer instead of the rounded answer, to say that there is a collision?

4) But what would be the conditions that must exist for the water to fall on the driver?
t(water)=1.5s
t(car)=1.8s
By a difference of 0.3s, I don't think the water will fall on the driver. To 2 significant digits, must t(water) be exactly equal to t(car) for the water to fall on the driver, theoretically?

3) The rule about rounding your final answer to the number of significant figures of the least well known value is a little over simplistic - 0.9 is actually a more accurate measurement than, say, 0.1, even though they both have the same number of significant digits. They both have an error of 0.05, which is about 5.6% of 0.9 and 50% of 0.1.

I'd do the calculation at full precision then allow for a 5.6% error (Rounding down to 1 s.f. is overdoing it). I think the intention is that you just use full precision and not worry about the errors involved, but its good that you worry about this kind of thing.

4) In real life the times wouldn't have to be exactly the same since both the driver and bucket have size - they're not just points. You haven't been given the dimensions of the bucket or the driver, so I wouldn't worry about this. In any case a car travelling at 24 km/h will advance about 2 metres in 0.3 seconds, so unless you have a 2 metre bucket or driver the guy should be safe.

Got it! Thanks!

5)

This is one of the 1D kinemantic equation:

displacement=(initial velocity)(delta t) + (1/2)(acceleration)(delta t)^2

and is solving for delta t

======================

I don't understand the circled parts! Why can you just omit the negative v2 and take the positive v2? And what is the reason of doing this?

(Note that this finally derives back to the equation acceleration = change in velocity / time)

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5) Does anyone know why?

Here's an (unimaginative) example: You throw a rock into the air, and watch it as it goes up, reaches a maximum height, then falls back down to the ground. You can model this motion with the equation d = ut + (1/2)a(t^2).

For heights below the maximum height, solving this equation for t will give TWO times because of the +/- before the square root, because the rock will reach this height twice: once on the way up, and once on the way back down.

The equation just assumes that the rock has been travelling like this since the beginning of space and time and flew up through the Earth's surface, which isn't what happened. This means that you can discard "nonsense" solutions, or times before the rock began it's epic journey. Does this help?

By the way, if your ultimate goal was to solve for "a," remember "a" also appears in the expression for V2. To get all the "a"s on one side of the equation will involve squaring V2, which will get rid of the +/- (so its not like you get two accelerations or anything).

That's not the fastest way to get the acceleration...

Where'd that nice scan go??????

przyk said:
Here's an (unimaginative) example: You throw a rock into the air, and watch it as it goes up, reaches a maximum height, then falls back down to the ground. You can model this motion with the equation d = ut + (1/2)a(t^2).

For heights below the maximum height, solving this equation for t will give TWO times because of the +/- before the square root, because the rock will reach this height twice: once on the way up, and once on the way back down.

The equation just assumes that the rock has been travelling like this since the beginning of space and time and flew up through the Earth's surface, which isn't what happened. This means that you can discard "nonsense" solutions, or times before the rock began it's epic journey. Does this help?

By the way, if your ultimate goal was to solve for "a," remember "a" also appears in the expression for V2. To get all the "a"s on one side of the equation will involve squaring V2, which will get rid of the +/- (so its not like you get two accelerations or anything).

That's not the fastest way to get the acceleration...

Why discard negative v2 and not positive v2?
Would delta t= (-v2 - v1) / a always give a value that is negative? If not, why should we discard this equation and use delta t= (v2 - v1) / a ?