A) Displacement after 6.0 s 43.2 m uphill.
B) Displacement after 9.0 s 43.2 m uphill.
Explanation:
A car moving upwards in a hill is [tex]12 ms^{-1}[/tex].
Its uniform backward acceleration is [tex]-1.6ms^{-2}[/tex]. (since backward acceleration is a negative acceleration, it is mentioned in negative)
We need to find the displacement of the car after some time.
Using the equation of the motion formula, we know can identify the displacement.
D=[tex]vt+\frac{1}{2} at^2[/tex].
a) Displacement after 6.0 seconds,
D = [tex]12(6.0)+\frac{1}{2}(-1.6)(6.0)^2[/tex].
=[tex]72+\frac{1}{2} (36)(-1.6).[/tex]
=[tex]72+\frac{1}{2}(-57.6).[/tex]
=72-28.8.
D=43.2 m.
b) Displacement after 9.0 seconds,
D= [tex]12(9.0)+\frac{1}{2}(-1.6)(9.0)^2[/tex].
=[tex]108+\frac{1}{2} (81)(-1.6).[/tex]
=[tex]108+\frac{1}{2}(-129.6).[/tex]
= 108-64.8.
D=43.2 m.
