Answer:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.
You plug your 1 into all spots where the variable is T. Then after you’ve done this, you simply simplify the problem. Then after simplifying you should have an equation where you are simply adding and subtracting which will give you the final answer of 2.
X=3T^2-12T+5
X=3(1)^2-12(1)+5
X=(3)^2-12+5
X=9-12+5
X=2
Answer:
<h2>1. Friction is A. a force</h2>
<h2>2. An unbalanced force is B. When the object moves and accelerates</h2>
Answer:
1.1299 x 10^-8 second
Explanation:
Period = 1 / f = 1 / (8.85 * 10^7) = 1.1299 x 01^-8 sec
Velocity is define as the time rate taken for a change in acceleration
