Answer:
V=20cm/s
Explanation:
The average speed is the distance total divided the time total:
[tex]V=X/T[/tex]
First stage:
T1=5s
[tex]v_{f} =v_{o} - at[/tex]
But, [tex]v_{f} =0[/tex] (decelerates to rest)
then: [tex]a =v_{o} /t=0.3/5=0.06m/s^{2}[/tex]
on the other hand:
[tex]x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m[/tex]
X1=75cm
Second stage:
T2=5s
[tex]x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m[/tex]
X2=125cm
Finally:
X=X1+X2=200cm
T=T1+T2=10s
V=X/T=20cm/s
The average speed of the particle over the whole time interval is 20 cm/s.
Given the following data:
To determine the average speed over the whole time interval:
First of all, we would determine the acceleration of the particle in scenario A.
[tex]V = U - at\\\\0 = 30 - a(5)\\\\0=30-5a\\\\5a = 30\\\\a =\frac{30}{5} \\\\a =6\;cm/s^2[/tex]
Next, we would find the distance covered by the particle in scenario A.
[tex]S = ut - \frac{1}{2} at^2\\\\S = 30(5) - \frac{1}{2} \times 6 \times 5^2\\\\S = 150-3\times 25\\\\S=150-75[/tex]
Distance, S = 75 cm.
For scenario B, we would find the distance covered by the particle:
[tex]X = ut + \frac{1}{2} at^2\\\\X = 0(5) + \frac{1}{2} \times 10 \times 5^2\\\\X = 5 \times 25[/tex]
X = 125 cm.
Now, we can determine the average speed over the whole time interval:
[tex]Total \;distance = distance \;A +distance\;B\\\\Total \;distance = 75 +125[/tex]
Total distance = 200 cm
[tex]Total \;time = time \;A +time\;B\\\\Total \;distance = 5 +5[/tex]
Total time = 10 seconds.
Mathematically, average speed is given by the formula:
[tex]Average\;speed = \frac{Total\;distance}{Total\;time} \\\\Average\;speed = \frac{200}{10}[/tex]
Average speed = 20 cm/s.
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