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Sagot :
ANSWER:
a)
b) 28 m/s
c) 60 sec
d) -2.8 m/s^2
e) 21 m/s
STEP-BY-STEP EXPLANATION:
For t = 0 s to t = 20 s
a = 1.4 m/s ^2
u = 0 m/s
Therefore:
[tex]\begin{gathered} v=u+at \\ \\ \text{ We replacing} \\ \\ v=0+1.4\cdot20 \\ \\ v=28\text{ m/s} \end{gathered}[/tex](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:
[tex]\begin{gathered} 1260=\frac{1}{2}(t+30)\cdot28 \\ t+30=1260\cdot\frac{2}{28} \\ t=90-30 \\ t=60\text{ s} \end{gathered}[/tex]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:
[tex]\begin{gathered} 0=28+a\cdot10 \\ \\ a=\frac{-28}{10} \\ \\ a=-2.8\text{ m/s}^2 \end{gathered}[/tex](e)
The average speed is the distance traveled in the total time, therefore:
[tex]\begin{gathered} V=\frac{d}{t} \\ \\ \text{ We replacing} \\ \\ V=\frac{1260}{60} \\ \\ V=21\text{ m/s} \end{gathered}[/tex]
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