Answer:
a) Maximum speed = 25.28 m/s
b) Total time = 27.27 s
c) Total distance traveled = 402.43 m
Explanation:
a) Maximum speed is obtained after the end of acceleration
v = u + at
v = 13.5 + 1.9 x 6.2 = 25.28 m/s
Maximum speed = 25.28 m/s
b) We have maximum speed = 25.28 m/s, then it decelerates 1.2 m/s² until it stops.
v = u + at
0 = 25.28 - 1.2 t
t = 21.07 s
Total time = 6.2 + 21.07 = 27.27 s
c) Distance traveled for the first 6.2 s
s = ut + 0.5 at²
s = 13.5 x 6.2 + 0.5 x 1.9 x 6.2² = 120.22 m
Distance traveled for the second 21.07 s
s = ut + 0.5 at²
s = 25.28 x 21.07 - 0.5 x 1.2 x 21.07² = 282.21 m
Total distance traveled = 120.22 + 282.21 = 402.43 m
Answer:
a) Maximum speed = 25.28 m/s
b) Total time = 27.26 s
c) Total distance traveled = 390,5537
Explanation:
In order to solve the first proble we just have to use the next formula:
[tex]Vf= Vo+ Acc-t\\Vf= 13,5 + 6,2*1,9\\Vf=25,28 m/s\\[/tex]
So the maximum speed would be 25,28 m/s.
THe total time of the trip will be given by adding the inital time plus the velocity divided by the acceleration rate:
[tex]Time= 6,2 s+ \frac{25,28}{-1,2} \\Time= 6,2 +21,06\\Time= 27,26[/tex]
Remember that when dealing with time in physics you will always use positive numbers since there is no negative time.
To calculate the total distance covered we use the next formula:
[tex]D=Vo*t+ 1/2a*t^2\\D= 13,5*6,2 + 1/2(1,9)(6,2)^2\\D=87,75+36,518\\D=124,268m[/tex]
This is the first part now we calculate it with the stopping of the car:
[tex]D=Vo*t+ 1/2a*t^2\\D=25,28*21,06+ 1/2(-1,2)(21,06)^2\\D=532,3968-266,1141\\D= 266,2857meters[/tex]
No we just add the two distances to discover the whole distance:
Total distance= 124,268+266,2857= 390,5537
