Answer:
Mathew will take 5 seconds to ride down the hill.
Explanation:
Given equation is:
![d(t)= v_(0)t +(1)/(2)at^2](https://img.qammunity.org/2019/formulas/mathematics/high-school/igwxa162ctz3nldt1i7kqyoyq1npla50r7.png)
Matthew is cycling of a speed of 4 meters/second. So,
![v_(0)= 4 m/s](https://img.qammunity.org/2019/formulas/mathematics/high-school/4j04lx3o0ulahb91ih6jobh06amijyiubr.png)
When he starts down a hill, the bike accelerates at a rate of 0.4 m/s². So,
![a= 0.4 m/s^2](https://img.qammunity.org/2019/formulas/mathematics/high-school/n59nbm640emyu7d7hrnqi1r76bkb9bodny.png)
The vertical distance from the top of the hill to the bottom of the hill is 25 meters. So,
meters.
Plugging the values into the above equation, we will get......
![25= 4t+(1)/(2)(0.4)t^2\\ \\ 25= 4t+0.2t^2\\ \\ 0.2t^2+4t-25=0](https://img.qammunity.org/2019/formulas/mathematics/high-school/e2izbsvaqt6yk5zq51wqooelu0kg7g4bqo.png)
Using quadratic formula, we will get.......
![t= (-4\pm √(4^2-4(0.2)(-25)))/(2(0.2))\\ \\ t=(-4\pm √(16+20))/(0.4)\\ \\ t=(-4\pm √(36))/(0.4)=(-4\pm 6)/(0.4)\\ \\ t=(-4+6)/(0.4)=(2)/(0.4)=5\\ \\or\\ \\ t=(-4-6)/(0.4)=(-10)/(0.4)=-25](https://img.qammunity.org/2019/formulas/mathematics/high-school/gonei7u1j6elpykcn0pa80gwtcfo6yefnr.png)
(Negative value is ignored as the time can't be negative)
So, Mathew will take 5 seconds to ride down the hill.