Answer:
The bear's acceleration is [tex]2\frac{m}{s^{2}}[/tex]
Explanation:
The magnitude of the acceleration can be calculated using the following equation :
[tex]a=\frac{dv}{dt}[/tex] (I)
Where ''a'' is the acceleration
Where ''dv'' is the speed variation
Where ''dt'' is the time variation
In this exercise, the time variation is equal to 3 seconds because it is the amount of time in which the bear accelerated from [tex]1.5\frac{m}{s}[/tex] to [tex]7.5\frac{m}{s}[/tex]
The speed variation is equal to :
[tex]dv=vf-vi[/tex]
Where ''vf'' is the final speed and ''vi'' is the initial speed.
Finally, we can calculate the bear's acceleration using the equation (I) :
[tex]a=\frac{dv}{dt}[/tex]
[tex]a=\frac{7.5\frac{m}{s}-1.5\frac{m}{s}}{3s}[/tex]
[tex]a=\frac{6\frac{m}{s}}{3s}[/tex]
[tex]a=2\frac{m}{s^{2}}[/tex]
We find that the bear's acceleration is [tex]2\frac{m}{s^{2}}[/tex]
