Answer:
- There was a constant acceleration at 0 to 10s
- There was a zero acceleration at 10 to 25s
- There was a constant deceleration at 25 to 30s
Explanation:
See attachment for complete question.
Solving (a): What happens at 0s to 10s
There was a constant acceleration and this is proven below.
At time 0, velocity = 15
At time 10, velocity = 30
This is represented as:
[tex](t_1,v_1) = (0,15)[/tex]
[tex](t_2,v_2) = (10,30)[/tex]
Acceleration (A) is the rate of change of velocity against time.
So:
[tex]A = \frac{v_2 - v_1}{t_2-t_1}[/tex]
[tex]A = \frac{30-15}{10 - 0}[/tex]
[tex]A = \frac{15}{10}[/tex]
[tex]A = 1.5[/tex]
Since the acceleration is positive, then it shows a constant acceleration.
Solving (b): What happens at 10s to 25s
There was a zero acceleration and this is because the velocity do not change.
See proof below
At time 10, velocity = 30
At time 25, velocity = 30
This is represented as:
[tex](t_1,v_1) = (10,30)[/tex]
[tex](t_2,v_2) = (25,30)[/tex]
Acceleration (A) is the rate of change of velocity against time.
So:
[tex]A = \frac{v_2 - v_1}{t_2-t_1}[/tex]
[tex]A = \frac{30-30}{25 - 10}[/tex]
[tex]A = \frac{0}{15}[/tex]
[tex]A = 0[/tex]
Solving (c): What happens at 25s to 30s
There was a constant deceleration and this is proven below.
At time 25, velocity = 30
At time 30, velocity = 0
This is represented as:
[tex](t_1,v_1) = (25,30)[/tex]
[tex](t_2,v_2) = (30,0)[/tex]
Acceleration (A) is the rate of change of velocity against time.
So:
[tex]A = \frac{v_2 - v_1}{t_2-t_1}[/tex]
[tex]A = \frac{0-30}{30-25}[/tex]
[tex]A = \frac{-30}{5}[/tex]
[tex]A = -6[/tex]
Since the acceleration is negative, then it shows a constant deceleration
