Answer:
Honda won by 0.14 s
Explanation:
We are given that
Distance =S=200 m
Initial velocity of Honda=u=0m/s
Initial velocity of Porsche=u'=0m/s
Acceleration of Honda=[tex]3.0m/s^2[/tex]
Acceleration of Porsche's=[tex]3.5m/s^2[/tex]
Time taken by Honda to start=1 s
[tex]s=ut+\frac{1}{2}at^2[/tex]
Substitute the values
[tex]200=0(t)+\frac{1}{2}(3)t^2[/tex]
[tex]200=\frac{3}{2}t^2[/tex]
[tex]t^2=\frac{200\times 2}{3}=\frac{400}{3}[/tex]
[tex]t=\sqrt{\frac{400}{3}}=11.55s [/tex]
Time taken by Honda=11.55 s
Now, time taken by Porsche
[tex]200=\frac{1}{2}(3.5)t^2[/tex]
[tex]t^2=\frac{200\times 2}{3.5}[/tex]
[tex]t=\sqrt{\frac{400}{3.5}}=10.69 s[/tex]
Total time taken by Porsche=10.69+1=11.69 s
Because it start 1 s late
Time taken by Honda is less than Porsche .Therefore, Honda won and
Time =11.69-11.55=0.14 s
Honda won by 0.14 s
The Honda will finish the race first by 0.1 s.
The given parameters:
Let the initial speed of each car = 0
The time of motion of the cars is calculated as follows;
[tex]s = v_0t + \frac{1}{2} at^2\\\\ s = 0 + \frac{1}{2} at^2\\\\ s = \frac{1}{2} at^2\\\\ t = \sqrt{\frac{2s}{a} } \\\\ [/tex]
The time for Porsche;
[tex]t_p = \sqrt{\frac{2s}{a} } \\\\ t_p = \sqrt{\frac{2 \times 200}{3.5} } \\\\ t_p = 10.7 \ s[/tex]
The time for Honda;
[tex]t_H = \sqrt{\frac{2s}{a} } - 1 \ s\\\\ t_H = \sqrt{\frac{2\times 200}{3} }\ - 1\ s\\\\ t_H = 10.6 \ s[/tex]
Thus, the Honda will finish the race first by 0.1 s.
Learn more about time of motion here: https://brainly.com/question/2364404
