Urgent Kinematics Help!

  • Thread starter Lucy_x
  • Start date
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Homework Statement



A car that is accelerating uniformly travels 18 m in the fifth second of its motion.
Calculate acceleration.

I've tried to answer this a million different ways.
If it's in the fifth second on motion that means it has travelled for four seconds, yes?
And I'm not sure that this question gives enough variables for it to be answered... would you agree?

Homework Equations



d=ut+0.5(at^2)


The Attempt at a Solution



The answer in my book says 4.0 is correct. I can't seem to get this.
d=ut+0.5(at^2)
18=0+0.5(16a)
36=16a
36/16
=2.25

Then i was thinking perhaps it has travelled 18 m in 0.5 seconds only, so i did
:d=ut+0.5(at^2)
18=0=0.5(0.25a)
36=0.25
36/0.25
=144

Please help!
 

Answers and Replies

  • #2
78
0

Homework Statement



A car that is accelerating uniformly travels 18 m in the fifth second of its motion.
Calculate acceleration.

I've tried to answer this a million different ways.
If it's in the fifth second on motion that means it has travelled for four seconds, yes?
And I'm not sure that this question gives enough variables for it to be answered... would you agree?

Homework Equations



d=ut+0.5(at^2)


The Attempt at a Solution



The answer in my book says 4.0 is correct. I can't seem to get this.
d=ut+0.5(at^2)
18=0+0.5(16a)
36=16a
36/16
=2.25

Then i was thinking perhaps it has travelled 18 m in 0.5 seconds only, so i did
:d=ut+0.5(at^2)
18=0=0.5(0.25a)
36=0.25
36/0.25
=144

Please help!

Haha that's a trickyyyyyy problem. First off, You were right in that they failed to mention that the initial velocity is 0. I guess they want you to assume that although it's almost against all scientific principles to assume lol. Secondly, you'll have to use calculus. Think about integrating acceleration, letting it just be constant. Then, use your answer from there as the velocity function, v(t), and take a definite integral. Remember that a definite integral does not bring about another constant of integration. Finally, "in the fifth second" can be considered poor wording as most people wouldn't read it as "the interval between the 4th and 5th second," which is what it actually means. Hope that helps.
 
  • #3
2
0
Thanks, just a few moments ago i realised that i can figure it out by calculating s1 and s2. Thanks for the answer, it helped a lot, though.
 

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