Answer:
a.After [tex]15[/tex] second Mr Comer's speed [tex]=[/tex] [tex]1.7 m/s[/tex]
b.Distance travelled by Mr.Comer in [tex]15[/tex] seconds [tex]=[/tex] [tex]18.75\ m[/tex]
Explanation:
a. Lets recall our first equation of motion [tex]V_{f} =V_{i} + at[/tex]
Now we know that [tex]V_{i} = 0.8m/s[/tex] , [tex]a = 0.06\ m/s^2[/tex] and
[tex]t = 15 \ sec[/tex]
Plugging the values we have.
[tex]V_{f} =V_{i} + at[/tex]
[tex]V_{f} =0.8 + 0.06 \times 15[/tex]
[tex]V_{f} =0.8+ 0.9[/tex]
[tex]V_{f} =1.7 m/s[/tex]
Then Mr.Comer's speed after [tex]15[/tex] sec [tex]=[/tex] [tex]1.7ms^-1[/tex]
b.
Lets find the distance and recall our third equation of motion.
[tex]V_{f} ^2-V{i}^2 = 2as[/tex]
So [tex]s =[/tex] distance covered.
Dividing both sides with 2a we have.
[tex]\frac{V_{f} ^2-V{i}^2}{2a} = 2as[/tex]
Plugging the values.
[tex]\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as[/tex]
[tex]s= 18.75\ m[/tex]
So Mr.Comer will travel a distance of [tex]s= 18.75\ m[/tex].
