Answer:
7) [tex]a=8m/s^2[/tex]
8) [tex]a=-5m/s^2[/tex]
9) [tex]a=-1.5m/s^2[/tex]
10) [tex]v=29.4m/s[/tex]
Explanation:
For the problems 7, 8 and 9 we just apply the definition of acceleration, since no more information is given, which is:
[tex]a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t_f-t_i}[/tex]
So for each problem we will have:
7) [tex]a=\frac{26m/s-10m/s}{2s}=8m/s^2[/tex]
8) [tex]a=\frac{10m/s-25m/s}{3s}=-5m/s^2[/tex]
9) [tex]a=\frac{0m/s-15m/s}{10s}=-1.5m/s^2[/tex]
For the problem 10, we use the equation of velocity in accelerated motion:
[tex]v=v_0+at[/tex]
Since the ball starts from rest and the acceleration is that of gravity (we take the downward direction positive), we have:
[tex]v=(0m/s)+(9.8m/s)(3s)=29.4m/s[/tex]