ANSWER:
a)
b) 28 m/s
c) 60 sec
d) -2.8 m/s^2
e) 21 m/s
Explanation:
For t = 0 s to t = 20 s
a = 1.4 m/s ^2
u = 0 m/s
Therefore:
![\begin{gathered} v=u+at \\ \\ \text{ We replacing} \\ \\ v=0+1.4\cdot20 \\ \\ v=28\text{ m/s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/vxg8o76k1usobvy656zql1ggyd9cacrv13.png)
(a)
Knowing this, we can sketch the velocity-time graph:
(b)
The maximum speed can be determined in the graph and is 28 m/s
(c)
Area under v-t graph gives displacement:
![\begin{gathered} 1260=(1)/(2)(t+30)\cdot28 \\ t+30=1260\cdot(2)/(28) \\ t=90-30 \\ t=60\text{ s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/kuz16hnqf48v5bocttogd570fx6lhvgujw.png)
So the total time is 60 seconds
(d)
Using between t = 50 and t = 60
t = 60 - 50 = 10 sec
u = 28 m/s
v = 0
We solving for a:
![\begin{gathered} 0=28+a\cdot10 \\ \\ a=(-28)/(10) \\ \\ a=-2.8\text{ m/s}^2 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/difhqklbhf2kkrvblwz89wp7piwvtcl440.png)
(e)
The average speed is the distance traveled in the total time, therefore:
![\begin{gathered} V=(d)/(t) \\ \\ \text{ We replacing} \\ \\ V=(1260)/(60) \\ \\ V=21\text{ m/s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/4d2q0qwi9s9396uhmlm2lio9jow004hb8p.png)