Answer:
[tex]a_{avg} = 1.17 m/s^{2}[/tex]
Given:
initial velocity, u = 0
final velocity, v = 33.7 m/s
t = 10 s
final velocity, v' = 17.6 m/s
t' = 5 s
Total time, T = 10 + 5 = 15 s
Solution:
The rate of change of velocity of an object is referred to as the acceleration of that object.
Average accelaeration, [tex]a_{avg} = \frac{\Deta v}{\Delta t}[/tex]
Now,
Initial acceleration of the body, [tex]a = \frac{v - u}{t}[/tex]
[tex]a = \frac{33.7 - 0}{10} = 3.37 m/s^{2}[/tex]
Now, average acceleration during the 15 seconds:
[tex]a = \frac{v' - u}{T}[/tex]
[tex]a = \frac{17.6 - 0}{15} = 1.17 m/s^{2}[/tex]
