Physics Help Forum displacement problem

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Jul 8th 2017, 05:05 PM #1 Junior Member   Join Date: Jun 2017 Posts: 15 displacement problem Hi, I got a different answer than the lesson text I am working on, and was hoping someone could explain. V2 = 31.1m/s a = 2.7m/s^2 t = 3.2s d = V2(t) - 1/2(a)(t^2) = (31.1m/s)(3.2s) - 1/2(2.7m/s^2)(3.2s)^2 = (99.6) - (1.35)(10.24) = 99.6 - 13.824 = 85.776m I understand it's a bit messy, and some calculations may be a few decimals off. The problem I am having is the lesson text I have says the answer is 113m (rounded I assume), where did I go wrong to not get that answer? Or is the lesson text wrong? Any help is appreciated, thanks very much.
 Jul 8th 2017, 08:24 PM #2 Senior Member   Join Date: Apr 2017 Posts: 361 You have subtracted the two numbers ,not sure why.... Adding them 99.6+13.824 gives the answer of 113 ... But you haven't supplied the full question so I can't give a definitive answer.
Jul 8th 2017, 10:33 PM   #3
Junior Member

Join Date: Jun 2017
Posts: 15
 Originally Posted by oz93666 You have subtracted the two numbers ,not sure why.... Adding them 99.6+13.824 gives the answer of 113 ... But you haven't supplied the full question so I can't give a definitive answer.

The full question is:

Find the displacement of a vehicle that achieves a velocity of 112km/h [W], accelerating at 2.7m/s^2 [W] for 3.2 seconds.

The lesson text uses the equation:
d = V2(t) - 1/2(a)(t^2)

and they get the answer 113m [W].

If that is correct, how do I get that number? I understand it's a simple minus and plus issue, but how come the minus doesn't seem to apply? Or am I missing something?

Thanks again, I appreciate your help.

 Jul 9th 2017, 12:22 AM #4 Senior Member   Join Date: Apr 2017 Posts: 361 so the final velocity is 31.1m/sec and it reaches this after accelerating for 3,2 secs at 2.7m/secsec v=at 3.2x2.7 = 8.64 m/sec This is the change in velocity after accelerating for 3.2 secs So we have an initial velocity of 31.1- 8.64 = 22.46 and final velocity 31.1 average velocity over the period of acceleration= 26.46 26.46 x 3.2 = 85.6 this is the displacement during acceleration We need the full question to resolve this .... without one word left out ... If you've given the exact full question already then there is an error in the book. Last edited by oz93666; Jul 9th 2017 at 12:26 AM.
Jul 9th 2017, 09:57 AM   #5
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Join Date: Jun 2017
Posts: 15
 Originally Posted by oz93666 so the final velocity is 31.1m/sec and it reaches this after accelerating for 3,2 secs at 2.7m/secsec v=at 3.2x2.7 = 8.64 m/sec This is the change in velocity after accelerating for 3.2 secs So we have an initial velocity of 31.1- 8.64 = 22.46 and final velocity 31.1 average velocity over the period of acceleration= 26.46 26.46 x 3.2 = 85.6 this is the displacement during acceleration We need the full question to resolve this .... without one word left out ... If you've given the exact full question already then there is an error in the book.

I uploaded a picture of the text to imgur, sorry if the reply with quote is unnecessary, still new to this.

http://imgur.com/Ar1JtSv

 Jul 9th 2017, 11:51 AM #6 Senior Member     Join Date: Aug 2008 Posts: 113 their equation is ok ... their arithmetic is incorrect. $\Delta x = v_f \cdot t - \dfrac{1}{2}at^2$ $\Delta x = (31.2 \, m/s)(3.2 \, s) - \dfrac{1}{2}(2.7 \, m/s^2)(3.2 \, s)^2 \ne 113 \, m$ $85.7 \, m$ is correct.
 Jul 9th 2017, 01:05 PM #7 Junior Member   Join Date: Jun 2017 Posts: 15 great thanks very much

 Tags acceleration, displacement, problem, speed, time, velocity

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