Before the driver applies the brakes ( with the reaction time ):
d 1 = v0 · t = 20 m/s · 0.53 s = 10.6 m
After that:
v = v0 - a · t1
0 = 20 m/s - 7 · t1
7 · t1 = 20
t1 = 2.86 s
d 2 = v 0 · t1 - a · t1² / 2
d 2 = 20 m/s · 2.86 s - 7 m/s² · (2.86 s)²/2 = 57.2 m - 28.6 m = 28.6 m
d = d 1 + d 2 = 10.6 m + 28.6 m = 39.2 m
Answer: the stopping distance of a car is 39.2 m.
Answer:
a i think
Step-by-step explanation:
Answer:
24
Step-by-step explanation:
If she did all 3of them today then you subtract a day from each one then it would look like this
5+4+15=24
Answer: You can multiply the top equation by -1 to eliminate the x variable.
And the solution is (2,4/3) in case you need it.
Step-by-step explanation:
2x + 3y = 8
2x + 6y = 12
If you multiply the upper equation or down equation by one, you will be able to eliminate the x variable.
-1( 2x + 3y) = -1(8) New equation: -2x -3y = -8.
Add the new equation you got by multiplying the top equation by -1 to the bottom equation.
Add them: -2x -3y = -8
2x + 6y = 12
3y = 4
y = 4/3
You can now input the value for y into the one of the equations and solve for x.
-2x - 3(4/3) = -8
-2x -4 = -8
+4 +4
-2x = -4
x = 2
Answer:
Option C 7 is your answer ☺️☺️☺️
