V = t^2 - 9t + 18
position, s
s = t^3 /3 - 4.5t^2 +18t + C
t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1
Average velocity: distance / time
distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
Average velocity = 27.67 / 8 = 3.46 m/s
t = 5 s
v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
speed = |-2| m/s = 2 m/s
Moving right
V > 0 => t^2 - 9t + 18 > 0
(t - 6)(t - 3) > 0
=> t > 6 and t > 3 => t > 6 s => Interval (6,8)
=> t < 6 and t <3 => t <3 s => interval (0,3)
Going faster and slowing dowm
acceleration, a = v' = 2t - 9
a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)
The answer is B tell me if I am wrong.
Answer:
1. W = F d = 20 N * 6 m = 120 J
2. F = W / d = 60 J / 2 m = 30 N
3. d = W / F = 350 J / 85 N = 4.12 m
4. P = W / t = F d / t = 45 N * 9 m / 10 s = 40.5 Watts
5. W = P t = 500 W * 120 sec = 60,000 J
6. t = W / P = 550 J / 310 W = 1.77 sec
The acceleration is the principal subordinate of the speed if the speed is steady the subsidiary is invalid if the speed is diminishing the subsidiary is negative. When discussing so much stuff we consider the momentary esteem.
<span>Note that when you back off, you back off by and large yet can locally in time quicken a tiny bit, suppose amid 1/tenth of a sec since you achieved a segment of the street which was slanting. In any case, this does not change the way that when the speed diminishes, the quickening is negative.</span>
Answer:
The speed is 15 km/h or 4.16 m/s.
Explanation:
A boat travels the distance that separates Gran Canaria from Tenerife (90 km) in 6 hours. Which the speed of the boat in km / h? And in m / s?
Given that,
Distance, d = 90 km = 90000 m
Time, t = 6 hours = 21600 s
Speed = distance/time
or
So, the required speed is 15 km/h or 4.16 m/s.
