Someone me with cal

a car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 44 ft/s2. what is the distance covered before the car comes to a stop? (give your answers correct to one decimal place.)

Since we are given a deceleration with a unit of ft/s2, it is convenient to convert 50 mi/h to ft/s. Appropriate conversion factors can be seen in the internet or in books. 

50 mi/h = 73.33 ft/s

the equation for the distance travelled is

x = Vo*t - 0.5*a*t^2

where 

x = distance travelled
Vo = initial velocity
t = time of travel
a = deceleration

since the time of travel is not known, we can find this using the equation

V = Vo - at

where V is the final velocity equal to zero (comes to stop)
substiruting values we get t = 1.67 s

therefore 

x = 61.1 ft (0.01 mi)

Look, when you want to do something like put it like x^2. x2 could mean x² or x₂. so to clear confusion, know the notation. to find the product multiply the x's and you get -8x and the y's and you get 6y. since it's all being multiplied it is just -8x6y, or answer a. although the better answer is -48xy(combining -8 and 6) because you should multiply constants whenever you can.