Answer:
Mathew will take 5 seconds to ride down the hill.
Step-by-step explanation:
Given equation is: [tex]d(t)= v_{0}t +\frac{1}{2}at^2[/tex]
Matthew is cycling of a speed of 4 meters/second. So, [tex]v_{0}= 4 m/s[/tex]
When he starts down a hill, the bike accelerates at a rate of 0.4 m/s². So, [tex]a= 0.4 m/s^2[/tex]
The vertical distance from the top of the hill to the bottom of the hill is 25 meters. So, [tex]d(t)= 25[/tex] meters.
Plugging the values into the above equation, we will get......
[tex]25= 4t+\frac{1}{2}(0.4)t^2\\ \\ 25= 4t+0.2t^2\\ \\ 0.2t^2+4t-25=0[/tex]
Using quadratic formula, we will get.......
[tex]t= \frac{-4\pm \sqrt{4^2-4(0.2)(-25)}}{2(0.2)}\\ \\ t=\frac{-4\pm \sqrt{16+20}}{0.4}\\ \\ t=\frac{-4\pm \sqrt{36}}{0.4}=\frac{-4\pm 6}{0.4}\\ \\ t=\frac{-4+6}{0.4}=\frac{2}{0.4}=5\\ \\or\\ \\ t=\frac{-4-6}{0.4}=\frac{-10}{0.4}=-25[/tex]
(Negative value is ignored as the time can't be negative)
So, Mathew will take 5 seconds to ride down the hill.
Answer:
It’ll only take Mathew 5 seconds to ride his bike down the hill.
Step-by-step explanation:
