Kinematics 1 dimensional motion acceleration question

  • Thread starter Hemingway
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  • #1
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A race driver has made a pit stop to refuel. After refueling, he starts from rest and leaves the pit area with an acceleration whose magnitude is 6.8 m/s2; after 4.1 s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant velocity of 66.0 m/s overtakes and passes the entering car. The entering car maintains its acceleration. How much time is required for the entering car to catch up with the other car?

b.2 Equations used
v2=u2+2as
s=ut+1/2at2
and other equations


The Attempt at a Solution



distance = v2=u2+2as
s= v2 - u2/2a
s= 263.1m

The answer is t = 11.2s
....

I am really stuck guys. I understand the final velocity of the entering car must be greater to catch up to the car travelling constant acceleration, but the displacement is the same. But I am finding it difficult to proceed without knowing the final velocity of entering car or the displacement.

Can someone offer a worked example for me - with different values?
 

Answers and Replies

  • #2
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Try breaking up the problem into two parts:
1- accelerating up to the race track edge, right when the passing vehicle is aside the car.
2- accelerating up to the vehicle ahead, to when the car catches up.

Hint: What distance has the accelerating car traveled when it's at the track edge? Can this be used to find the velocity of the accelerating car at that moment? You are right about the displacement of the two cars being the same when they've caught up, but it is from this point, not the start of the pit-stop car's acceleration.
 
  • #3
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Try breaking up the problem into two parts:
1- accelerating up to the race track edge, right when the passing vehicle is aside the car.
2- accelerating up to the vehicle ahead, to when the car catches up.

Hint: What distance has the accelerating car traveled when it's at the track edge? Can this be used to find the velocity of the accelerating car at that moment? You are right about the displacement of the two cars being the same when they've caught up, but it is from this point, not the start of the pit-stop car's acceleration.
Thanks for the suggestion, I might have to wait for my lecturer to get back to me, I am just burnt out on this problem. I just don't know how to work out the displacement when I don't know the accelerating car's final velocity when overtaking the car, and cannot seem to understand..

My assumption that entering car's final velocity isn't equal to 66m/s is the point where I lose the problem..
dang.
 
  • #4
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Don't give up. Try this:
d = ut + 1/2at^2
where u = 0m/s, a = 6.8m/s^2, t = 4.1 s
Calculate d
Then try using:
v^2 = u^2 + 2ad
Then go from there.

Get back to me with your solution.
 
Last edited:
  • #5
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Don't give up. Try this:
d = ut + 1/2at^2
where u = 0m/s, a = 6.8m/s^2, t = 4.1 s
Calculate d
Then try using:
v^2 = u^2 + 2ad
Then go from there.

Get back to me with your solution.
1. d = 57.15m
2. v = 27.87m/s

Thanks, I understand where I went wrong with D now. I am still just very daft with how to find out velocity of the car when it catches up with the other car. I know it must be some assumption that I'm missing. thanks so much for your help!
 
  • #6
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Excellent. You're almost there! Keep going!
Now you can equate the displacements of each car (like you said in your first post, "the displacement is the same") from that point and solve for the time it takes them to reach the distance at which they catch up.
Hint: Da = Vat and Db = Ubt + 1/2at^2
where Da = displacement of car 'a', Va = velocity of car 'a', and similar for car 'b'
and Da = Db so...

I should be more clear, car 'a' is the one passing at this moment. Car 'b' is the pit-stop car. Hope I didn't confuse you.
 
Last edited:
  • #7
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Excellent. You're almost there! Keep going!
Now you can equate the displacements of each car from that point and solve for the time it takes them to reach that distance.
Hint: Da = Vat and Db = Ubt + 1/2at^2
where Da = displacement of car 'a', Va = velocity of car 'a', and similar for car 'b'
and Da = Db so...

I should be more clear, car 'a' is the one passing at this moment. Car 'b' is the pit-stop car. Hope I didn't confuse you.
D = VT
= 57.15/66.0
= 0.9s

D= ut + 1/2 at2
57.15 = 0t + 1/2 (6.8) (t2)
t2 = 114.3/6.8 = 16.8
t = 4.1s

oops that looks familiar

D= ut + 1/2 at2
57.15 = 27.87t + 1/2 at2

is this the one I should use... but that looks weird too

the answer said t = 11sec

So sorry :/ ....
 
  • #8
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Don't be sorry, you're so close. You're doing a good job. I'll give you a little nudge.

Da = Vat = 66t
Db = 27.87t + 1/2(6.8)t^2 = 27.87t + 3.4t^2
Now
Da = Db, plug in for each what we have above
66t = 27.87t + 3.4t^2
Do you know how to solve this for t?

Remember that the first time the question gave was for the pit-stop car to travel to the race track (t = 4.1s was for the 'first' part of the question), not to catch up. Now that we know the speed of each car at the moment they are side by side, and the acceleration of car 'b', we can calcluate the time t it takes the cars to meet up from the moment the one passes the other.
 
  • #9
42
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Don't be sorry, you're so close. You're doing a good job. I'll give you a little nudge.

Da = Vat = 66t
Db = 27.87t + 1/2(6.8)t^2 = 27.87t + 3.4t^2
Now
Da = Db, plug in for each what we have above
66t = 27.87t + 3.4t^2
Do you know how to solve this for t?

I thought I did but remembered I cant make it = 2t2, as I can't add them. Ah yes, I see what I did with Da, silly me.

You don't have to explain about finding t, you've already spent so much time on me! if you could send me a link or something that'd be wonderful :)
 
  • #10
49
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I'm enjoying helping you. And you're so close already. Let's continue, shall we? You're just a hop, skip, and a jump away from the answer.

First thing, subtract 27.87t from both sides of the equation. What do you get?
 
  • #11
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I'm enjoying teaching you this. And you're so close already. Let's continue, shall we? You're just a hop, skip, and a jump away from the answer.

First thing, subtract 27.87t from both sides. What do you get?
66t/28.87 = 2.368

will ts cancel as they are both associated so,

66t-27.87t = 3.4t^2
38.13 = 3.4t^2
t^2 = 11.21 !!!!


YYYYYAAAAAAYYYYYYYYYYYY - THANK YOU SSSOOOOO MUCH!!!!!!!!
 
  • #12
42
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66t/28.87 = 2.368

will ts cancel as they are both associated so,

66t-27.87t = 3.4t^2
38.13 = 3.4t^2
t^2 = 11.21 !!!!


YYYYYAAAAAAYYYYYYYYYYYY - THANK YOU SSSOOOOO MUCH!!!!!!!!
and then hang on.... might be too soon

t^2 = 11.21
t = 3.35
 
  • #13
49
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will ts cancel as they are both associated so,

66t-27.87t = 3.4t^2
38.13 = 3.4t^2
t^2 = 11.21 !!!!
I think it's just a typo, since you understood to cancel the common t, but it should read:

38.13 = 3.4t
11.2 s = t

YYYYYAAAAAAYYYYYYYYYYYY - THANK YOU SSSOOOOO MUCH!!!!!!!!
You're welcome. Your interest and tenacity got you the right answer. Kudos.

Oh, and btw. Did you notice how you didn't even need the final velocity of the pit-stop car to ascertain the time asked for in the question?

Happy studying.
 
Last edited:
  • #14
42
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Thanks again - you really, really helped my confidence :)
 

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