anonymous
  • anonymous
A 1.2 x 10^3 kg car accelerates uniformly from 5.0 m/s east to 12 m/s east. During this acceleration the car travels 94 m. What is the net force acting on the car during the acceleration? I know F=mass x acceleartion. We have the mass so I need to find the acceleration first. I know acceleration is velocity over time. So 12 m/s - 5 m/s = 7 m/s. We need the time, and I know that time is distance over velocity, so 94m / 7m/s = 13s. Now I can try to find the acceleration. 7m/s / 13s = 0.54 m/s^2. So force = 1200 kg x 0.54 m/s^2. I get 648 N, but the answer is 760 N. What am I doing wrong
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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DanJS
  • DanJS
do you remember what the kinematic equations are
DanJS
  • DanJS
there is a trick in figuring those out, where the time variable falls out of one of those equations, in terms of other variables
DanJS
  • DanJS
when you are deriving them

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DanJS
  • DanJS
you remember this looking thing? combo of the others \[Vf^2=Vo^2+2a*(Xf-Xo)\]
lochana
  • lochana
" so 94m / 7m/s = 13s" <-- this is not correct
lochana
  • lochana
@DanJS please continue:)
DanJS
  • DanJS
you can solve directly from that.. given the change in velocity, and the change in distance, you can figure the acceleration... (constant)
DanJS
  • DanJS
Accelerated from 5 to 12 m/s over a distance of 94m
DanJS
  • DanJS
remember, the acceleration is a CHANGE in a velocity over a time , you put 7 m/s, the change in the velocity , in to the distance formula d=v*t, in the distance formula, that is a constant velocity value, not the change in 2 values
DanJS
  • DanJS
oh, i was waiting for you to respond @lochana , i just now realized this is not your original question.. ... hahah
lochana
  • lochana
okay. so we can use \[V^2 = U^2 + 2as \] for this problem right? http://www.schoolphysics.co.uk/age14-16/Mechanics/Motion/text/Equations_of_motion/index.html
DanJS
  • DanJS
remember the first or second week of phys 1, you started with newton's laws i think to derive these... for constant value for acceleration https://images.search.yahoo.com/images/view;_ylt=AwrCxCsEGURWLgoAQwBXNyoA;_ylu=X3oDMTBsY2xzZWFmBGNvbG8DYmYxBHBvcwMxBHNlYwNzYw--?p=kinematic%20equation&back=https%3A%2F%2Fsearch.yahoo.com%2Fsearch%3Ffr%3Dyfp-t-901-s%26fr2%3Dsb-top-search%26hspart%3Dyahoo%26hsimp%3Dyhs-default%26p%3Dkinematic%2520equation&sigb=13ma6f0i6&imgurl=physicsclub.net%2Fimages%2FkinematicEquations.gif&sigi=11dprknmo&rurl=http%3A%2F%2Fphysicsclub.net%2FphysletIndex%2FlinearKinematics.html&sigr=11p5s0g07&name=Kinematic%20Equations&sign=10jn6dsgt&tt=Kinematic%20Equations&sigt=10jn6dsgt&no=7&w=300&h=179&size=2KB
lochana
  • lochana
u = initial velocity which is 7m/s v = last velocity which is 12m/s s = distance that is 94 a = acceleration
lochana
  • lochana
yes. that's right
lochana
  • lochana
so \[a = \frac{V^2 - V0^2}{2(x-x0)}\]\[a = \frac{144 - 25}{2*94}\]
DanJS
  • DanJS
a constant accel , means you can write the acceleration definition a=change in V / Change in Time as the first kinematic equation a*t = Vf - Vi
DanJS
  • DanJS
since a is taken as a constant
lochana
  • lochana
yes. but our goal is to find 'a'. we still don't know about 't' (time).
lochana
  • lochana
\[a = \frac{144 -25}{2*94} = \frac{119}{188}ms^{-2}\] this is the constant acceleration. m is given (1.2 x 10^3 kg)
DanJS
  • DanJS
if acceleration is constant, the velocity changes at a uniform rate... remember the average velocity.. =change in velocity over 2
lochana
  • lochana
\[F_{net} = ma\]
lochana
  • lochana
@DanJS do you think my solution for 'a' is correct?
DanJS
  • DanJS
THe average velocity is also the total distance over the total time = (Xf - Xi)/(tf - ti)
lochana
  • lochana
can you show steps to find "a"?
DanJS
  • DanJS
combine those two , and you get the next kinematic thing to remember, \[= \frac{ \Delta x }{ \Delta t } = \frac{ \Delta V }{ 2 }\]
DanJS
  • DanJS
i found what I was typing.. Follow this, see how you can figure out that kinematic equation relating Velocity, Distance, and Acceleration
DanJS
  • DanJS
There are other ways, this is straight forward from other ideas of accel and velocity and position http://theory.uwinnipeg.ca/physics/onedim/node7.html#eq:_first_EOM
lochana
  • lochana
@Openstudystudent314 I didn't notice that you have given the solution to your question. look at my answer. you get a = 119/188. now multiply that with 1200kg. you get 760N.
lochana
  • lochana
@DanJS yes. what you are saying is correct. we can solve this question in many ways.
DanJS
  • DanJS
i was trying to show how you can derive that equation i gave, in the process it shows the time variable dropping out, and you can relate accel to distance and change in velocity
DanJS
  • DanJS
constant accel, uniform change in velocity
lochana
  • lochana
yeah. I remember that. it takes some extra steps to solve it. otherwise both are same:)
DanJS
  • DanJS
i think we used the Force = dp/dt to start, not sure.. long time back
DanJS
  • DanJS
change in momentum

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