Answer:
1. Â 3 [tex]m/s^{2}[/tex]
2. Â 1.5 [tex]m/s^{2}[/tex]
3. Â 3 seconds
4. Â 0 [tex]m/s^{2}[/tex]
5. Â 2.2 seconds
Explanation:
(1)
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making a the subject we have
[tex]a=\frac {v-u}{t}[/tex]
Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds
a = [tex]\frac {30-0}{10}=3 m/s^{2}[/tex]
(2)
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making a the subject we have
[tex]a=\frac {v-u}{t}[/tex]
Substituting u=10m/s, v=22m/s and t=8 seconds
a = [tex]\frac {22-10}{8}=1.5 m/s^{2}[/tex]
(3)
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making t the subject we have
[tex]t=\frac {v-u}{a}[/tex]
Substituting u=0m/s since at rest, v=15m/s and a=5 [tex]\frac {m}{s^{2}}[/tex]
= [tex]\frac {15-0}{5}=3s[/tex]
(4)
When initial and final velocity are constant, there’s no acceleration as proven below
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making a the subject we have
[tex]a=\frac {v-u}{t}[/tex]
Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds
a = [tex]\frac {20-20}{10}=0 m/s^{2}[/tex]
(5)
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making t the subject we have
[tex]t=\frac {v-u}{a}[/tex]
Substituting u=9m/s since at rest, v=0m/s and a=-4.1 [tex]\frac {m}{s^{2}}[/tex]
= [tex]\frac {0-9}{-4.1}=2.2s[/tex]
