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3 years ago

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A car going initially with a velocity 13.5 m/s accelerates at a rate of 1.9 m/s for 6.2 s. It then accelerates at a rate of-1.2

m/s until it stops. a)Find the car's maximum speed. b) Find the total time from the start of the first acceleration until the car is stopped. c) What is the total distance the car travelled?

2 answers:

gregori [183]3 years ago

6 0

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:

$Vf= Vo+ Acc-t\\Vf= 13,5 + 6,2*1,9\\Vf=25,28 m/s\\$

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:

$Time= 6,2 s+ \frac{25,28}{-1,2} \\Time= 6,2 +21,06\\Time= 27,26$

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:

$D=Vo*t+ 1/2a*t^2\\D= 13,5*6,2 + 1/2(1,9)(6,2)^2\\D=87,75+36,518\\D=124,268m$

This is the first part now we calculate it with the stopping of the car:

$D=Vo*t+ 1/2a*t^2\\D=25,28*21,06+ 1/2(-1,2)(21,06)^2\\D=532,3968-266,1141\\D= 266,2857meters$

No we just add the two distances to discover the whole distance:

Total distance= 124,268+266,2857= 390,5537

yanalaym [24]3 years ago

5 0

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

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The critical velocity for a satellite orbiting around a planet is given by

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$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

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The period of the orbit is given by the circumference of the orbit divided by the velocity:

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$R=6.37\cdot 10^6 m$

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Substituting,

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