Answer:
6.31 mi
Step-by-step explanation:
The diagram below explains the solution better.
From the diagram,
C = starting point of the race.
A = end of the first part of the race.
B = end of the race.
Using Cosine rule, we can find the straight-line distance between the starting point and the end of the race.
Cosine rule states that:
![a^2 = b^2 + c^2 - 2bc[cos(A)]](https://tex.z-dn.net/?f=a%5E2%20%3D%20b%5E2%20%2B%20c%5E2%20-%202bc%5Bcos%28A%29%5D)
where A = angle A = <A
Given that
b = 5.2 miles
c = 2.0 miles
<A = 115° (from the diagram)
Hence,
![a^2 = 5.2^2 + 2.0^2 - 2*5.2*2.0[cos(115)]\\\\a^2 = 27.04 + 4 - 20.8[cos(115)]\\\\a^2 = 31.04 + 8.79\\\\a^2 = 39.83\\\\a = \sqrt{39.83}\\ \\a = 6.31 mi](https://tex.z-dn.net/?f=a%5E2%20%3D%205.2%5E2%20%2B%202.0%5E2%20-%202%2A5.2%2A2.0%5Bcos%28115%29%5D%5C%5C%5C%5Ca%5E2%20%3D%2027.04%20%2B%204%20-%2020.8%5Bcos%28115%29%5D%5C%5C%5C%5Ca%5E2%20%3D%2031.04%20%2B%208.79%5C%5C%5C%5Ca%5E2%20%3D%2039.83%5C%5C%5C%5Ca%20%3D%20%5Csqrt%7B39.83%7D%5C%5C%20%5C%5Ca%20%3D%206.31%20mi)
The straight-line distance between the starting point and the end of the race is 6.31 mi
Answer:
3- 8 -2
Step-by-step explanation:
HOPE THIS HELPS
Answer:
3 sips equal a gulp
Step-by-step explanation:
we have
----> equation A
Divide by 2 both sides
----> equation B
equate equation A and equation B
Group terms that contain the same variable
therefore
3 sips equal a gulp
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
