When you see a traffic light turn red, you apply the brakes until you come to a stop. Suppose your initial speed was 12.9 m/s, and you come to rest in 35.0 m. How much time does this take? Assume constant deceleration.

What equation do I use to solve this problem? You can show me some steps to solve the problem. Thanks

Well, buckle up, my friend, because we're about to solve this problem in style! To figure out the time it takes for you to come to a stop, we can use the kinematic equation:

vf^2 = vi^2 + 2ad,

Where vf is the final velocity (which is 0 since you come to a stop), vi is the initial velocity (12.9 m/s in this case), a is the acceleration (which is negative since you are decelerating), and d is the distance (35.0 m).

Here's how we can solve it step by step:

Step 1: Plug in the given values into the equation:
0^2 = (12.9 m/s)^2 + 2a(35.0 m).

Step 2: Simplify the equation further:
0 = 166.41 m^2/s^2 + 2a(35.0 m).

Step 3: Rearrange the equation to isolate the acceleration term:
2a(35.0 m) = -166.41 m^2/s^2.

Step 4: Divide both sides of the equation by 70.0 m (2 times 35.0 m):
a = -166.41 m^2/s^2 ÷ (2 × 35.0 m).

Step 5: Calculate the value of a:
a = -4.754 m/s^2 (approximately).

Step 6: Now, we can use another kinematic equation to find the time (t) it takes to come to a stop:
vf = vi + at.

Considering that vf is 0 m/s, vi is 12.9 m/s, and a is -4.754 m/s^2, we get:
0 = 12.9 m/s + (-4.754 m/s^2) × t.

Step 7: Rearrange the equation to isolate the time term:
4.754 m/s^2 × t = 12.9 m/s.

Step 8: Divide both sides by 4.754 m/s^2 to solve for t:
t ≈ 2.71 s.

Ta-dah! According to my calculations, it takes approximately 2.71 seconds for you to come to a stop. Just make sure to leave enough time for everyone else to appreciate your stylish braking maneuver!

To solve this problem, you can use the following equation of motion:

v^2 = u^2 + 2aS

where:
v = final velocity (0 m/s, as the car comes to rest)
u = initial velocity (12.9 m/s)
a = acceleration (which is the negative value of deceleration)
S = distance (35.0 m)

Step 1: Rearrange the equation to solve for acceleration:
a = (v^2 - u^2)/(2S)

Step 2: Substitute the given values into the equation:
a = (0^2 - (12.9)^2)/(2(35.0))

Step 3: Calculate:
a = (-166.41)/70.0
≈ -2.37 m/s^2

Step 4: Now, we can calculate the time using another equation of motion:

v = u + at

Since the final velocity (v) is 0, we can rearrange the equation to solve for time (t):

t = (v - u) / a

Step 5: Substituting the values into the equation:

t = (0 - 12.9) / (-2.37)
= 12.9 / 2.37
≈ 5.44 seconds

Therefore, it will take approximately 5.44 seconds for the car to come to a stop.

To solve this problem, we can use the equations of motion with constant acceleration. In this case, the initial speed is 12.9 m/s, the final speed is 0 m/s (since you come to a stop), and the distance is given as 35.0 m.

The equation we can use is:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s)
u = initial velocity (12.9 m/s)
a = acceleration (which we need to calculate)
s = distance (35.0 m)

We can rearrange the equation to solve for acceleration:

a = (v^2 - u^2) / (2s)

Plugging in the values, we get:

a = (0^2 - 12.9^2) / (2 * 35.0)

a = (-166.41) / 70.0

a ≈ -2.3771 m/s^2 (note the negative sign indicating deceleration)

Now, to find the time it takes to come to a stop, we can use the equation:

v = u + at

Since the final velocity is 0 m/s, we can rearrange the equation to solve for time:

t = (v - u) / a

Plugging in the values, we have:

t = (0 - 12.9) / (-2.3771)

t ≈ 5.43 seconds

Therefore, it takes approximately 5.43 seconds to come to a stop.

V^2 = Vo + 2ad.

V = 0.
Vo = 12.9 m/s.
d = 35 m.
a = ?. Solve for a. It will be negative.

V = Vo + at.
V = 0.
Vo = 12.9 m/s.
a = Previous calculated value.
t = ?.