It’s B lovey !!! Glad to help
If the track has circumference c miles, then and the car's speeds are x and y mi/s, then since time = distance/speed,
<span>In the same direction, the slow car has a "lead" of 1 mile which the faster car has to make up. </span>
<span>1/(y-x) = 120 </span>
<span>1/(x+y) = 30 </span>
<span>120(y-x) = 30(x+y) </span>
<span>120y-120x = 30x+30y </span>
<span>90y = 150x </span>
<span>y = 5/3 x </span>
<span>1/(x + 5/3 x) = 30 </span>
<span>8/3 x * 30 = 1 </span>
<span>80x = 1 </span>
<span>x = 1/80 mi/sec = 45 mph </span>
<span>y = 75 mph</span>
Answer:
can u send a picture to identify it
Explanation:
sorry I just need point:/
Answer: Your question is missing below is the question
Question : What is the no-friction needed speed (in m/s ) for these turns?
answer:
20.1 m/s
Explanation:
2.5 mile track
number of turns = 4
length of each turn = 0.25 mile
banked at 9 12'
<u>Determine the no-friction needed speed </u>
First step : calculate the value of R
2πR / 4 = πR / 2
note : πR / 2 = 0.25 mile
∴ R = ( 0.25 * 2 ) / π
= 0.159 mile ≈ 256 m
Finally no-friction needed speed
tan θ = v^2 / gR
∴ v^2 = gR * tan θ
v = √9.81 * 256 * tan(9.2°) = 20.1 m/s