We appreciate your visit to When Jim and Rob ride bicycles Jim can only accelerate at three quarters the acceleration of Rob Both start from rest at the bottom of. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
The time has taken that should Jim start in order to reach the top at the same time as Rob is 46 s.
Calculation of the time taken
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
Learn more about time here; https://brainly.com/question/24727111
Thanks for taking the time to read When Jim and Rob ride bicycles Jim can only accelerate at three quarters the acceleration of Rob Both start from rest at the bottom of. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
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
