Answer:
[tex]0.3 m/s^2[/tex] is Bob's average acceleration.
Explanation:
Initial velocity of the bob's in-lineskates= u = 0 m/s
Acceleration of the Bob= a = ?
Final velocity of the bob's in-lineskates= v = 4.5 m/s
Duration of time in which 0 m/s velocity is changed to 4.5 m/s = t = 15 s
Using first equation of motion: v = u + at
[tex]4.5 m/s=0m/s+a\times (15 s)[/tex]
[tex]5.0 m/s-0m/s =a\times 15 s[/tex]
[tex]4.5 m/s=a\times 15 s[/tex]
[tex]a=\frac{4.5 m/s}{15 s}=0.3 m/s^2[/tex]
[tex]0.3 m/s^2[/tex] is Bob's average acceleration.
