Answer:
The distance is 58.03 m
Explanation:
Constant Acceleration Motion
It occurs when the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:
[tex]v_f=v_o+at[/tex]
The distance traveled by the object is given by:
[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}[/tex]
The conditions of the problem state the cyclist has an initial speed of v0=20.7 m/s during t=7.8 seconds and acceleration of -3.4 m/s^2.
The final speed is:
[tex]v_f=20.7+(-3.4)\cdot 7.8[/tex]
[tex]v_f=20.7-26.52[/tex]
[tex]v_f=-5.82\ m/s[/tex]
Note the cyclist has stopped and come back because his speed is negative. Now calculate the distance:
[tex]\displaystyle x=20.7\cdot 7.8+\frac{(-3.4)\cdot 7.8^2}{2}[/tex]
[tex]\displaystyle x=161.46-103.43[/tex]
x=58.03 m
