Answer:
48m
Explanation:
Given the following data;
Initial velocity = 0m/s
Final velocity = 6m/s
Time, t = 6 secs
Time, T2 = 5 secs
Mathematically, acceleration is given by the equation;
![Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}](https://tex.z-dn.net/?f=Acceleration%20%28a%29%20%3D%20%5Cfrac%7Bfinal%20%5C%3B%20velocity%20%20-%20%20initial%20%5C%3B%20velocity%7D%7Btime%7D)
Substituting into the equation;
Acceleration, a = 1m/s²
<u>To find the distance covered in the first phase;</u>
<em>Solving for distance, we would use the second equation of motion;</em>
![S = ut + \frac {1}{2}at^{2}](https://tex.z-dn.net/?f=%20S%20%3D%20ut%20%2B%20%5Cfrac%20%7B1%7D%7B2%7Dat%5E%7B2%7D)
<em>Substituting the values into the equation;</em>
![S = 0(6) + \frac {1}{2}*1*(6)^{2}](https://tex.z-dn.net/?f=%20S%20%3D%200%286%29%20%2B%20%5Cfrac%20%7B1%7D%7B2%7D%2A1%2A%286%29%5E%7B2%7D)
![S = 0 + \frac {1}{2}*1*36](https://tex.z-dn.net/?f=%20S%20%3D%200%20%2B%20%5Cfrac%20%7B1%7D%7B2%7D%2A1%2A36)
![S = 0.5 *36](https://tex.z-dn.net/?f=%20S%20%3D%200.5%20%2A36)
Distance, S1 = 18m
<u>For the second phase, time T2 = 5 secs;</u>
<em>Mathematically, speed is given by the equation;</em>
![Speed = \frac{distance}{time}](https://tex.z-dn.net/?f=Speed%20%3D%20%5Cfrac%7Bdistance%7D%7Btime%7D)
<em>Making distance the subject of formula, we have;</em>
![Distance, S = speed * time](https://tex.z-dn.net/?f=Distance%2C%20S%20%3D%20speed%20%2A%20time)
<em>Substituting into the above equation;</em>
![Distance, S = 6 * 5](https://tex.z-dn.net/?f=Distance%2C%20S%20%3D%206%20%2A%205)
Distance, S2 = 30m
Total distance = S1 + S2 = 18m + 30m = 48m
Total distance = 48m
<em>Therefore, the total distance traveled by the biker is 48m.</em>