Answer:the car was traveling at a speed of 80 ft/s when the brakes were first applied.
Step-by-step explanation:
The car braked with a constant deceleration of 16ft/s^2. This is a negative acceleration. Therefore,
a = - 16ft/s^2
While decelerating, the car produced skid marks measuring 200 feet before coming to a stop.
This means that it travelled a distance,
s = 200 feet
We want to determine how fast the car was traveling (in ft/s) when the brakes were first applied. This is the car's initial velocity, u.
Since the car came to a stop, its final velocity, v = 0
Applying Newton's equation of motion,
v^2 = u^2 + 2as
0 = u^2 - 2 × 16 × 200
u^2 = 6400
u = √6400
u = 80 ft/s
Answer:
x= negative 1 and 2/5 y=1 and 3/5
Step-by-step explanation:
1. Start by putting both equations into slope-intercept form. (y=mx+b)
-3x+3y=9
-3x+3x+3y=9+3x
3y=9+3x
3y/3=9+3x/3
y=3+x
2x-7y=-14
2x-2x-7y=-14-2x
-7y=-14-2x
-7y/-7=-14/-7-2x/-7
y=2+2/7x
2. Graph the equations and find where they intercept.
Distance=speed x time
so it would be 28 x 6=168
It would be 32-4x=12
32-4(5)=12
32-20=12
12=12
Sum means to Add.
So the sum of d and 9 would be d + 9
