Question

Interpreting Date: Bob straps on his

Answer 1

Answer:

$0.3 m/s^2$ 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

$4.5 m/s=0m/s+a\times (15 s)$

$5.0 m/s-0m/s =a\times 15 s$

$4.5 m/s=a\times 15 s$

$a=\frac{4.5 m/s}{15 s}=0.3 m/s^2$

$0.3 m/s^2$ is Bob's  average acceleration.