Physics Help Forum

Physics Help Forum (http://physicshelpforum.com/physics-help-forum.php)
-   Kinematics and Dynamics (http://physicshelpforum.com/kinematics-dynamics/)
-   -   displacement problem (http://physicshelpforum.com/kinematics-dynamics/13330-displacement-problem.html)

jpompey Jul 8th 2017 05:05 PM

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.

oz93666 Jul 8th 2017 08:24 PM

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.

jpompey Jul 8th 2017 10:33 PM

Quote:

Originally Posted by oz93666 (Post 35834)
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.

oz93666 Jul 9th 2017 12:22 AM

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.

jpompey Jul 9th 2017 09:57 AM

Quote:

Originally Posted by oz93666 (Post 35840)
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.

http://imgur.com/Ar1JtSv

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

skeeter Jul 9th 2017 11:51 AM

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.

jpompey Jul 9th 2017 01:05 PM

great thanks very much


All times are GMT -7. The time now is 09:50 PM.

Copyright © 2016 Physics Help Forum. All rights reserved.
Copyright © 2008-2012 Physics Help Forum. All rights reserved.