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

Question

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

danjs · Answer

do you remember what the kinematic equations are

danjs · Answer

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 · Answer

when you are deriving them

danjs · Answer

you remember this looking thing? combo of the others

$$Vf^2=Vo^2+2a*(Xf-Xo)$$

lochana · Answer

" so 94m / 7m/s = 13s" <-- this is not correct

lochana · Answer

@DanJS please continue:)

danjs · Answer

you can solve directly from that.. given the change in velocity, and the change in distance, you can figure the acceleration... (constant)

danjs · Answer

Accelerated from 5 to 12 m/s   over a distance of 94m

danjs · Answer

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 · Answer

oh, i was waiting for you to respond @lochana ,  i just now realized this is not your original question.. ... hahah

lochana · Answer

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 · Answer

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 · Answer

u = initial velocity which is 7m/s
v = last velocity which is 12m/s
s = distance that is 94

a = acceleration

lochana · Answer

yes. that's right

lochana · Answer

so $$a = \frac{V^2 - V0^2}{2(x-x0)}$$$$a = \frac{144 - 25}{2*94}$$

danjs · Answer

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 · Answer

since a is taken as a constant

lochana · Answer

yes. but our goal is to find 'a'. we still don't know about 't' (time).

lochana · Answer

$$a = \frac{144 -25}{2*94} = \frac{119}{188}ms^{-2}$$
this is the constant acceleration. 

m is given (1.2 x 10^3 kg)

danjs · Answer

if acceleration is constant, the velocity changes at a uniform rate... remember the average velocity.. =change in velocity over 2

lochana · Answer

$$F_{net} = ma$$

lochana · Answer

@DanJS do you think my solution for 'a' is correct?

danjs · Answer

THe average velocity is also the total distance over the total time  <V> = (Xf - Xi)/(tf - ti)

lochana · Answer

can you show steps to find "a"?

danjs · Answer

combine those two , and you get the next kinematic thing to remember, 

$$<V >= \frac{ \Delta x }{ \Delta t } = \frac{ \Delta V }{ 2 }$$

danjs · Answer

i found what I was typing.. 

Follow this,  see how you can figure out that kinematic equation relating Velocity, Distance, and Acceleration

danjs · Answer

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 · Answer

@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 · Answer

@DanJS yes. what you are saying is correct. we can solve this question in many ways.

danjs · Answer

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 · Answer

constant accel, uniform change in velocity

lochana · Answer

yeah. I remember that. it takes some extra steps to solve it. otherwise both are same:)

danjs · Answer

i think we used the Force = dp/dt  to start,  not sure.. long time back

danjs · Answer

change in momentum