Answer:
a = 1.428 [m/s²]
v₀ = 5 [m/s]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]x=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}[/tex]
where:
x = final point [m]
x₀ = initial point [m]
v₀ = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
But we need to use this additional equation.
[tex]v_{f}=v_{o}+a*t[/tex]
where:
vf = final velocity = 15 [m/s]
Now we can use this equation, replacing it, in the first one. We must bear in mind that the difference among x - x₀ is equal to 70 [m]
[tex]x-x_{o}=v_{o}*t+\frac{1}{2}*a*t^{2} \\x-x_{o}=(v_{f}-a*t)*t+\frac{1}{2} *a*t^{2}\\70=(15-a*t)*t+\frac{1}{2}*a*t^{2}\\70=15*t-a*t^{2} +\frac{1}{2}*a*t^{2} \\70=15*t-\frac{1}{2}*a*t^{2}\\70=15*(7)-\frac{1}{2} *a*(7)^{2}\\105-70=0.5*a*49\\35=24.5*a\\a=1.428[m/s^{2} ][/tex]
Now replacing this value in the second equation, we can find the initial velocity.
[tex]15=v_{o}+1.428*7\\v_{o}=5[m/s][/tex]
