# Thread: Did I do this right. F=ma to find Vi for my question.

1. ## Did I do this right. F=ma to find Vi for my question.

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

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
Code:
So I found acceleration of a=F/m = 2765/2600 = 1.06346m/s^2
that should be minus. -2765 and -1.06346m/s^2

s=vt-0.5*a*t^2

3036.3=0-0.5*-1.06346*t^2

3036.5/(0.5*1.06346)=t^2

t=75.56605s

Vi = 0 - (-1.06346)(75.56605) = 80.3614 m/s

3. Babymilo,

Not trying to step on your toes here, i just want to show an alternative.

$\frac{1}{2}mv^2$ = KE of vi

$\frac{1}{2}mv^2$ = work done by brakes

work done by brakes = braking force x distance.

vi = 80.36 m/s like you stated.