Explanation:
|F|= B/2
B/2 = √(A²+B²+2ABcosθ) ------(1)
Since the resultant of A and B is perpendicular to Vector A
tan90°= BSinθ/(A+BCosθ)
(A+BCosθ)=0
Cosθ=-A/B ----(2)
Using equation (1)
B/2 = √(A²+B²+2ABcosθ)
B/2 = √(A²+B²+2AB×-A/B)
B/2=√(A²+B²-2A²)
B/2=√(B²-A²)
B²/4=B²-A²
A²=B²-B²/4
A²=3B²/4
A=√3B/2
Using equation (2)
Cosθ=-A/B
Cosθ=-[√3B/2]/B
Cosθ=-√3/2
θ= cos^-1 (-√3/2)
θ= 150°
Answer:
The car stops in 7.78s and does not spare the child.
Explanation:
In order to know if the car stops before the distance to the child, you take into account the following equation:
(1)
vo: initial speed of the car = 45km/h
a: deceleration of the car = 2 m/s^2
t: time
xo: initial distance to the child = 25m
x: final distance to the child = 0m
It is necessary that the solution of the equation (1) for time t are real.
You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:
![45\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=12.5\frac{m}{s}](https://tex.z-dn.net/?f=45%5Cfrac%7Bkm%7D%7Bh%7D%2A%5Cfrac%7B1h%7D%7B3600s%7D%2A%5Cfrac%7B1000m%7D%7B1km%7D%3D12.5%5Cfrac%7Bm%7D%7Bs%7D)
![0=25+12.5t-2t^2\\\\2t^2-12.5t-25=0\\\\t_{1,2}=\frac{-(-12.5)\pm \sqrt{(-12.5)^2-4(2)(-25)}}{2(2)}\\\\t_{1,2}=\frac{12.25\pm 18.87}{4}\\\\t_1=7.78s\\\\t_2=-1.65s](https://tex.z-dn.net/?f=0%3D25%2B12.5t-2t%5E2%5C%5C%5C%5C2t%5E2-12.5t-25%3D0%5C%5C%5C%5Ct_%7B1%2C2%7D%3D%5Cfrac%7B-%28-12.5%29%5Cpm%20%5Csqrt%7B%28-12.5%29%5E2-4%282%29%28-25%29%7D%7D%7B2%282%29%7D%5C%5C%5C%5Ct_%7B1%2C2%7D%3D%5Cfrac%7B12.25%5Cpm%2018.87%7D%7B4%7D%5C%5C%5C%5Ct_1%3D7.78s%5C%5C%5C%5Ct_2%3D-1.65s)
You take the first value t1 because it has physical meaning.
The solution for t is real, then, the car stops in 7.78s and does not spare the child.
Answer:
D. The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.
Explanation:
I just got it right lol
Give me some answer choices and i will be happy to help
Answer: Option D
We know that speed is defined as the derivative of the position with respect to time. The derivative shows the slope changes that occur in the graph of the function shown.
Observe the attached image.
Notice that at the beginning the graph resembles a line whose slope is constant. This means that during this time interval (t1) the speed is approximately constant and greater than zero. Then we observe a time interval during which the function is approximately horizontal, parallel to the time axis. In this interval (t2) the slope is approximately 0. This means that in this interval the velocity is approximately 0. Finally, the graph has a positive and large slope. This means that during the last time interval (t3) the speed increases. Therefore the correct option is D.
D. The velocity decreases, then increases.