Did I do this right an F=ma need to find Initial Speed?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
5 replies · 2K views
bradycat
Messages
13
Reaction score
0
Did I do this right an F=ma need to find Initial Speed?

Need help on this.
A car whose mass is 2600 kg can produce an unbalanced braking force of 2765N.
Calculate the car's initial speed in order to have a stopping distance of 3036.3m.
Vi=?
So...
m=2600kg
F=2765N
s=3036.3m
Vf=0

So I found acceleration of a=F/m = 2765/2600 = 1.06346m/s^2
So I have to find time=Square root of 2(3036.33)/1.06346 = 75.56605s
So I would use this formula to solve for Vi? right
Vi=Vf -at
Vi = 0 - (1.06346)(75.56605) = -80.3614 m/s
Is the negative right?
IS THIS CORRECT if not please show me where I went wrong
Thanks
Jo
 
Physics news on Phys.org


Where you have found the time you have made a mistake since that formula requires the initial velocity too.
 


Kurdt said:
Where you have found the time you have made a mistake since that formula requires the initial velocity too.

Hi Kurdt,
I used the time formula where you use distancex2 over acceleration. THAT is wrong your saying?
What formula should I use?
 


Yes, that's wrong. It came from d=v0t + (1/2)at^2; you only get t=sqrt(2d/a) if v0=0, which, in this case, it's not. However, you got the right answer at the end because acceleration is symmetrical with respect to time. That is, if you record a car decelerating to 0 at -1.06 m/s^2 and play the video backwards at 1 second per second, the car would seem to be accelerating at 1.06 m/s^2 to its initial speed. That's why using t=sqrt(2d/a) worked.
 


Hi Ideasule.
So the -80.3614 m/s is correct?
So the Vi can have a negative number then?
I thought it could not have a negative on a Vi?
 


Velocities and accelerations are vector quantities, so they can certainly be negative. For Vi, yes, -80.36 m/s. If you don't explain the fact about time symmetry, however, you wouldn't get full credit on a test.

Whether Vi is positive or negative depends on which way you define as positive. Here, you used a=1.06 m/s^2, which means you've defined "backwards" as positive. It's not surprising, then, that initial velocity would be negative. (Initial SPEED, however, is not; speed is a scalar.)