Find the Acceleration of the Aar Going Downhill

[Trix]

Homework Statement

A car originally traveling at 30 m/s manages to break for 5 sec. while traveling 125 m downhill. At that point the breaks fail. After an additional 5 sec. it travels an additional 150 m down the hill. What was the acceleration of the car after the breaks failed?

Homework Equations

Vxf = Vxi + axt and solve for a,Xf - Xi = Vxit + (1/2)axt^2, or ax = (Vxf - Vxi)/t

The Attempt at a Solution

Well, I've tried to use all 3 equations to no avail, perhaps I'm using the wrong ones or I don't understand the question. Or maybe it wants the instantaneous acceleration at that instant (when breaks fail)? Any tips, suggestions are welcome, thanks.

 

[Redbelly98]

You'll need to apply those equations to two separate pieces of the car's motion, (1) with breaks on the first 5 sec and (2) with breaks off for the next 5 sec.

 

[thomate1]

First, let's define some variables:
Vxi = initial velocity of the car (30 m/s)
Vxf = final velocity of the car (unknown)
ax = acceleration of the car after the brakes fail (unknown)
t = time after the brakes fail (5 seconds)

Using the equation Vxf = Vxi + axt, we can solve for the final velocity of the car:
Vxf = Vxi + axt
Vxf = 30 m/s + ax(5 s)
Vxf = 30 m/s + 5ax

Next, we can use the equation Xf - Xi = Vxit + (1/2)axt^2 to find the distance traveled by the car after the brakes fail:
Xf - Xi = (30 m/s)(5 s) + (1/2)ax(5 s)^2
Xf - Xi = 150 m + 12.5ax

Now, we know that the total distance traveled by the car after the brakes fail is 150 m + 125 m = 275 m. So we can set Xf - Xi = 275 m and solve for ax:
275 m = 150 m + 12.5ax
125 m = 12.5ax
ax = 10 m/s^2

Therefore, the acceleration of the car after the brakes fail is 10 m/s^2. This is the instantaneous acceleration at the moment when the brakes fail.