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2 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]2 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]2 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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Answer:

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Explanation:

From the question,

μk = a/g...................... Equation 1

Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.

make a the subject of the equation

a = μk(g).................. Equation 2

Given: μk = 0.160, g = 9.8 m/s²

Substitute into equation 2

a = 0.16(9.8)

a = 1.568 m/s²

Using,

F = ma

Where F = force, m = mass.

Make m the subject of the equation

m = F/a................... Equation 3

m = 160/1.568

m = 102.04 kg.

Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.

Assuming the initial velocity of the skier to be zero

Fd+mgμ = 1/2mv²........................Equation 4

Where v = speed of the skier after crossing the patch, d = distance/width of the patch.

v = √2(Fd+mgμ)/m)................ Equation 5

Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16

Substitute these values into equation 5

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Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water

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To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

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Where,

$V_i =$ Velocity in each state

g= Gravity

h = Height

Our values are given as,

$A_1 = 2.4*10^{-4} m^2$

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$h = 0.11m$

Replacing at the kinetic equation to find $V_2$ we have,

$v_2 = \sqrt{v_1^2 + 2gh}$

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$v_2= 1.67 m/s$

Applying the concepts of continuity,

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We need to find A_2 then,

$A_2= \frac{A_1v_1 }{v_2}$

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Answer:

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<u>A</u><u>nswer:</u>

<u>The </u><u>magnitude</u><u> </u><u>of </u><u>its </u><u>acceleration</u><u> </u><u>is </u><u>1</u><u>.</u><u>6</u><u> </u><u>m/</u><u>s^</u><u>2</u><u>.</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

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