Answer: [tex]15 m/s^{2}[/tex]
Explanation:
We have the following data:
[tex]d=180 m[/tex] is the distance traveled by the car
[tex]t= 6 s[/tex] is the time of travel
[tex]u=45 m/s[/tex] is the car's final velocity
[tex]v[/tex] is the unknown car's initial velocity
[tex]a[/tex] is the car's constant acceleration
Now, we can solve this problem with the following equations of motion:
[tex]u=v+at[/tex] (1)
[tex]d=vt+\frac{1}{2}at^{2}[/tex] (2)
Isolating [tex]v[/tex] from (1):
[tex]v=u-at[/tex] (3)
[tex]v=45 m/s-(6 s)a[/tex] (4)
Substituting (4) in (2):
[tex]d=(45 m/s-(6 s)a)t+\frac{1}{2}at^{2}[/tex] (5)
Solving and finding [tex]a[/tex]:
[tex]180 m=(45 m/s)(6 s)-36 s^{2}a+\frac{1}{2}a(6 s)^{2}[/tex] (6)
Finally:
[tex]a=15 m/s^{2}[/tex] This is the car's acceleration
