Answer:
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, \sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}∑F=
r
1
mv
2
=
r
2
mu
2
................ii
form i and ii we can write
v^2= \frac{1}{2} u^2v
2
=
2
1
u
2
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v
Here's a list that might help:
Odd functions: sinx, tanx, cotx, cscx
Even functions: cosx, secx
We know that time is distance over velocity, that is:
time = distance / velocity
t = d / v
Now from the given:
d / 460 + d / 500 = 4.8
Now we can find for d:
d = 4.8 / [1/460 + 1/500]
d = 1150 miles
So the time of each flight is:
flight = 1150 / 460 = 2.5 hours
<span>return flight = 1150 / 500 = 2.3 hours</span>
