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Sagot :
Total distance = 36500 m
The average velocity = 19.73 m/s
Further explanation
Given
vo=initial velocity=0(from rest)
a=acceleration= 1 m/s²
t₁ = 20 s
t₂ = 0.5 hr = 1800 s
t₃= 30 s
Required
Total distance
Solution
State 1 : acceleration
[tex]\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m[/tex]
[tex]\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s[/tex]
State 2 : constant speed
[tex]\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m[/tex]
State 3 : deceleration
[tex]\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)[/tex]
[tex]\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m[/tex]
Total distance : state 1+ state 2+state 3
[tex]\tt 200 + 36000 + 300=36500~m[/tex]
the average velocity = total distance : total time
[tex]\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s[/tex]
The total distance covered by the car will be 36500m and the average velocity of the car for the entire trip will be 19.73 m/sec.
What is average velocity?
The average velocity is the ratio of the total distance traveled to the total time taken by the body. Its unit is m/sec.
The given data in the problem is;
v₀ is the initial velocity=0(from rest)
a is the acceleration= 1 m/s²
t₁ is the time period 1= 20 s
t₂ is the time period 2= 0.5 hr = 1800 s
t₃ is the time period 3= 30 s
For part 1
The distance travelled is found as;
[tex]\rm d=v_0t+\frac{1}{2} at^2 \\\\ \rm d=\frac{1}{2}\times 1 \times (20)^2 \\\\ \rm d=200\ m[/tex]
The initial velocity is found as ;
[tex]\rm v_t=v_0+at\\\\ \rm v_t=0+1 \times 20\\\\ \rm v_t=20 \ m/sec[/tex]
For part 2
The distance traveled is found as;
[tex]\rm d= v \times t \\\\ \rm d= 20 \times 1800 \\\\ \rm d= 36000 \ m[/tex]
For part 3
The acceleration is found as;
[tex]\rm v_t=v_0+at\\\\ \rm 0=v_0+a \times 30\\\\ \rm 20=-30 a \\\\ a= - \frac{2}{3}\ m/sec^2[/tex]
The distance traveled is found as;
[tex]\rm d=v_0t+\frac{1}{2} at^2 \\\\ \rm d=20.30 +\frac{1}{2} \times \frac{2}{3} (30)^2 \\\\ \rm d=300 \ m[/tex]
Total distance travelled= distance travelled( part 1+ part 2+ part3)
Total distance travelled=200+3600+300 = 36500 m
The average velocity is found as;
[tex]v_{avg}= \frac{36500}{20+1800+30} \\\\ v_{avg}= 19.73 \ sec[/tex]
Hence the total distance covered by the car will be 36500m and the average velocity of the car for the entire trip will be 19.73 m/sec.
To learn more about the average velocity refer to the link;
https://brainly.com/question/862972
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