Answer:
This question assumes that the car accelerates at the same rate as when it went from 0 to 60km/h
24.29m/s or 87.4km/h
Explanation:
Let's find the acceleration of the car:
let vi=0, vf=60km/h (16.67m/s), Δt = 8.0s
a = (vf-vi)/Δt
a = (16.67m/s-0)/8.0
a = 2.08m/s^2
Now we can use this acceleration to find vf in the second part:
50km/h is 13.89m/s
a = (vf-vi)Δt
vf = aΔt + vi
vf = 2.08m/s^2*5.0+13.89m/s
vf = 24.29m/s (87.4km/h)
You can compare the velocity of the car, 60 mph, with the velocity that a mass would acquire when falls from certain height.
First, convert 60 mph to m/s:
60 miles/h * 1.60 km/mile * 1000 m/km * 1h/3600s = 26.67 m/s
Second, calculate from what height a body in free fall reachs 26.67 m/s velocity when hits the floor.
free fall => Vf^2 = 2g*H => H = Vf^2 / (2g)
H = (26.67m/s)^2 / (2*9.8 m/s) = 36.2 m
If you consider that the height between the floors of a building is approximately 3.6 m, you get 36.2 m / 3.6 m/floor = 10 floors.
Then, you conclude that the force of impact is the same as driving you vehicle off a 10 story building.
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Answer: 4.98 m/s
Explanation:
You solve these kinetic energy, potential energy problems by using the fact P.E.+ K.E. = a constant as long as friction is ignored.
PEi = 0 in this case
KEi = ½mVi² = PEf+KEf = mghf + ½mVf²
½1210*8.31² = 1210*9.8*2.26 + ½1210*Vf²
½1210*Vf² = ½1210*8.31² - 1210*9.8*2.26
Vf² = 8.31² - 2*9.8*2.26 = 4.98² so Vf = 4.98m/s
Answer: car B has travelled 4times as far as Car A
d=vi*t+1/2at^2
No initial velocity so equation becomes;
d=1/2at^2 and the acceleration is the same between both only time is different;
Car A d=1/2a(1)^2
Car B d=1/2a(2)^2
Car A d= 1^2=1
Car B d= 2^2=4
Car B d=4*Car A
So car B has travelled 4 times as far as car A
