Answer:
[tex]acceleration=0.8 \frac{m}{s^{2} } \\distance=112.5 m[/tex]
Explanation:
We can find these answers following the equations of motion.
[tex]a=[/tex]Δ[tex]V/t[/tex]
Where Δ[tex]V[/tex] is the difference between the final speed and the initial speed. And [tex]t[/tex] is the time spent
We replace the terms:
[tex]a=\frac{18\frac{m}{s} -12\frac{m}{s} }{7.5s}[/tex]
We solve the difference:
[tex]a=\frac{6\frac{m}{s}}{7.5s}[/tex]
We divide the terms, so we can have the answer:
[tex]a=0.8 \frac{m}{s^{2}}[/tex]
2. To find the distance traveled by the truck, we use the equation:
[tex]x=V_0t+\frac{1}{2}at^{2}[/tex]
Where [tex]x[/tex] is the distance traveled, [tex]V_0[/tex] is the initial speed, [tex]a[/tex] is the acceleration and [tex]t[/tex] is the time.
We replace the terms:
[tex]x=(12\frac{m}{s}*7.5s)+\frac{1}{2}[0.8\frac{m}{s^{2} }*(7.5)^{2} ][/tex]
We multiply and solve the exponential:
[tex]x=90m+\frac{1}{2}(0.8\frac{m}{s^{2} }*56.25s^{2} )[/tex]
Then, we multiply the terms left:
[tex]x=90m+22.5m[/tex]
And add, so we can have the answer:
[tex]x=112.5m[/tex]
