At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Get detailed and accurate answers to your questions from a community of experts on our comprehensive Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
Answer:
[tex]D_T=18.567m[/tex]
Explanation:
From the question we are told that:
Acceleration [tex]a=8.0 m/s^2[/tex]
Displacement [tex]d=1.05 m[/tex]
Initial time [tex]t_1=6.0s[/tex]
Final Time [tex]t_2=2.5s[/tex]
Generally the equation for Velocity of 1.05 travel is mathematically given by
Using Newton's Law of Motion
[tex]V^2=2as[/tex]
[tex]V=\sqrt{2*6*1.05}[/tex]
[tex]V=4.1m/s[/tex]
Generally the equation for Distance traveled before stop is mathematically given by
[tex]d_2=v*t_1[/tex]
[tex]d_2=3.098*4[/tex]
[tex]d_2=12.392[/tex]
Generally the equation for Distance to stop is mathematically given by
Since For this Final section
Final velocity [tex]v_3=0 m/s[/tex]
Initial velocity [tex]u_3=4.1 m/s[/tex]
Therefore
Using Newton's Law of Motion
[tex]-a_3=(4.1)/(2.5)[/tex]
[tex]-a_3=1.64m/s^2[/tex]
Giving
[tex]v_3^2=u^2-2ad_3[/tex]
Therefore
[tex]d_3=\frac{u_3^2}{2ad_3}[/tex]
[tex]d_3=\frac{4.1^2}{2*1.64}[/tex]
[tex]d_3=5.125m[/tex]
Generally the Total Distance Traveled is mathematically given by
[tex]D_T=d_1+d_2+d_3[/tex]
[tex]D_T=5.125m+12.392+1.05 m[/tex]
[tex]D_T=18.567m[/tex]
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.