A box is given an initial leftward velocity of -20 m/s. Friction works against that motion, causing a constant acceleration of 4 m/s2 until the box stops. Hint: would acceleration be positive or negative here to slow the box down?

How far does the box travel before it comes to rest?
How much time does the box take to stop?

if v is - , a is +

v = Vi + a t
0 = -20 + 4 t
t = 5 seconds

x = Xi + Vi t + .5 a t^2
x = 0 - 20(5) + 2 (25)
x = -100 + 50
x = -50

I understand how you got the time but I dont really understand how x is -50

The box can travel in a negative?

sure, it has a negative velocity (leftward)

so it ends up at negative, not positive, x

To solve this problem, we can use the equations of motion to find the distance and time it takes for the box to come to a stop.

First, let's determine whether the acceleration will be positive or negative. Since the friction is working against the leftward motion of the box, the acceleration will be in the direction opposite to its initial velocity. Therefore, the acceleration will be negative.

Now let's find the time it takes for the box to stop. We can use the equation:

v = u + at

where:
v = final velocity (0 m/s because the box comes to a stop)
u = initial velocity (-20 m/s)
a = acceleration (-4 m/s^2)
t = time

Rearranging the equation to solve for t, we have:

t = (v - u) / a

Substituting the given values, we get:

t = (0 - (-20)) / (-4)
t = 20 / 4
t = 5 seconds

So, it takes 5 seconds for the box to come to a stop.

Next, let's find the distance the box travels before coming to a stop. We can use the equation:

s = ut + (1/2)at^2

where:
s = distance
u = initial velocity (-20 m/s)
t = time (5 seconds)
a = acceleration (-4 m/s^2)

Substituting the given values, we have:

s = (-20)(5) + (1/2)(-4)(5)^2
s = -100 + (-10)(25)
s = -100 - 250
s = -350 meters

Therefore, the box travels 350 meters in the leftward direction before it comes to rest. Note that the negative sign indicates the direction of motion, in this case, leftward.