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
H + 2 divided by 4
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Answer:
10x+2y
8c+2
0
Step-by-step explanation:
B = 26in, c = 40in
a^2 = 26^2 + 40^2
= 2276 (square root it)
a = 2root569 // 47.7in
Answer:
y = 2/3x +2
Step-by-step explanation:
First we need to find the slope
m = (y2-y1)/(x2-x1)
= (6-2)/(6-0)
=4/6
=2/3
The slope intercept form of the equation for a line is
y= mx+b
y =2/3 x +b
We know the y intercept is 2 (the y intercept is when x=0)
y = 2/3x +2
