Answer:
46.4 s
Explanation:
5 minutes = 60 * 5 = 300 seconds
Let g = 9.8 m/s2. And [tex]\theta[/tex] be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim . Both of their motions are subjected to parallel component of the gravitational acceleration [tex]gsin\theta[/tex]
Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]
Jim equation of motion is [tex]s = a_Jt_J^2/2 = (3a/4)t_J^2/2 = 3at_J^2/8[/tex]
As both of them cover the same distance
[tex] 45000a = 3at_J^2/8[/tex]
[tex]t_J^2 = 45000*8/3 = 120000[/tex]
[tex]t_J = \sqrt{120000} = 346.4 s[/tex]
So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time
The time has taken that should Jim start in order to reach the top at the same time as Rob is 46 s.
Since
x = v0t + 1/2 at^2
so x rob = 1/2 a 300^2
a rob = 2 x/300^2
a jim = 3/4*2*x/300^2
Now
x jim = 1/2*3/4*2*x/300^2*t^2
1 = 3/4*1/300^2 t^2
t^2 = 300^2*4/3
t = 2*300/sqrt(3)= 346 s
Now the time taken should be
= 346-300
= 46 s earlier
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