a) A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 40 ft/s^2. What is the distance covered before the car comes to a stop?

b) A car braked with a constant deceleration of 40 ft/s^2 and produced skid marks measuring 160 feet before coming to a stop. How fast was the car traveling when the brakes were first applied?

a) To find the distance covered before the car comes to a stop, we can use the equation of motion:

v^2 = u^2 + 2as

Where:
- v is the final velocity (0 since the car comes to a stop)
- u is the initial velocity (given as 50 mi/h)
- a is the acceleration (deceleration in this case, given as 40 ft/s^2)
- s is the distance covered (what we need to find)

First, let's convert the initial velocity from miles per hour to feet per second. Since 1 mile = 5280 feet and 1 hour = 3600 seconds, we can convert 50 mi/h to:

u = 50 mi/h * 5280 ft/mi * (1/3600) h/s ≈ 73.333 ft/s

Now, we can rearrange the equation to solve for s:

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

Substituting the known values:

s = (0^2 - (73.333 ft/s)^2) / (2 * -40 ft/s^2)
= (-5389.56 ft^2/s^2) / (-80 ft/s^2)
≈ 67.37 ft

Therefore, the distance covered before the car comes to a stop is approximately 67.37 feet.

b) To find the initial velocity of the car when the brakes were first applied, we can use the equation of motion mentioned above, but slightly rearranged:

v^2 = u^2 + 2as

This time, we know the distance (s) but need to find the initial velocity (u).

Rearranging the equation, we have:

u^2 = v^2 - 2as

Substituting the known values:

u^2 = (0 ft/s)^2 - 2 * (-40 ft/s^2) * 160 ft
= 0 - (-12800 ft^2/s^2)
= 12800 ft^2/s^2

Taking the square root of both sides to solve for u:

u = √(12800 ft^2/s^2)
≈ 113.14 ft/s

So, the car was traveling approximately 113.14 feet per second when the brakes were first applied.

a) To find the distance covered before the car comes to a stop, we can use the equations of motion. We know that the initial velocity (u) is 50 mi/h, the deceleration (a) is 40 ft/s^2, and we need to find the distance (s) covered.

First, let's convert the initial velocity from mi/h to ft/s:
1 mi = 5280 ft
1 h = 3600 s

Initial velocity (u) = 50 mi/h * 5280 ft/mi / 3600 s/h = 73.33 ft/s

Now we can use the equation of motion:
v^2 = u^2 + 2as

At the moment the car comes to a stop, the final velocity (v) is 0 ft/s. Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)

Plugging in the values:
s = (0^2 - 73.33^2) / (2 * -40)
s = (-5368.8889) / (-80)
s = 67.11 ft

Therefore, the distance covered before the car comes to a stop is 67.11 ft.

b) To find the initial velocity of the car when the brakes were first applied, we can use the equation of motion again. We know that the deceleration (a) is 40 ft/s^2, the distance (s) is 160 ft, and we need to find the initial velocity (u).

Again, we can use the equation of motion:
v^2 = u^2 + 2as

At the moment the car comes to a stop, the final velocity (v) is 0 ft/s. Rearranging the equation, we have:
u^2 = v^2 - 2as

Plugging in the values:
u^2 = 0^2 - 2 * 40 * 160
u^2 = -6400
There is no real solution for the initial velocity. The negative under the square root indicates that the given values do not result in a physically possible scenario.

Therefore, there is no valid initial velocity that corresponds to the given conditions.

a) Vo = 50 mph = 73.33 ft/s

Vo = sqrt (2 a X)
X = Vo^2/(2 a)

b) Use the Vo formula above, with X = 160 ft. The answer will be in ft/s.