Answer:
33 seconds.
Explanation:
The equation for speed with constant acceleration at time t its:
[tex]V(t) \ = \ V_0 \ + \ a \ t[/tex]
where [tex]V_0[/tex] is the initial speed, and a its the acceleration.
Starting at rest, the initial speed will be zero, so
[tex]V_0 = 0[/tex]
the final speed is
[tex]V(t_{f1}) = 23 \frac{m}{s}[/tex]
and the acceleration is
[tex]a = 2.3 \frac{m}{s^2}[/tex].
Taking all this together, we got
[tex]V(t_{f1}) = 23 \frac{m}{s} = 0 + 2.3 \ \frac{m}{s^2} t_{f1}[/tex]
[tex]23 \frac{m}{s} = 2.3 \ \frac{m}{s^2} t_{f1}[/tex]
[tex]\frac{23 \frac{m}{s}}{2.3 \ \frac{m}{s^2}} = t_{f1}[/tex]
[tex]10 s = t_{f1}[/tex]
So, for the first half of the problem we got a time of 10 seconds.
Now, the initial speed will be
[tex]V_0 = 23 \frac{m}{s}[/tex],
the acceleration
[tex]a=-1.0 \frac{m}{s^2}[/tex],
with a minus sign cause its slowing down, the final speed will be
[tex]V(t_{f2}) = 0[/tex]
Taking all together:
[tex]V(t_{f2}) = 0 = 23 \frac{m}{s} - 1.0 \frac{m}{s^2} t_{f2}[/tex]
[tex] 23 \frac{m}{s} = 1.0 \frac{m}{s^2} t_{f2}[/tex]
[tex] \frac{23 \frac{m}{s}}{1.0 \frac{m}{s^2}} = t_{f2}[/tex]
[tex] 23 s = t_{f2}[/tex]
So, for the first half of the problem we got a time of 23 seconds.
[tex]t_total = t_{f1} + t_{f2} = 33 \ s[/tex]
