1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
pantera1 [17]
2 years ago
13

A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10r, where a and fare in m/s^2

and seconds, respectively. After reaching a speed of 125 m/s, a parachute is deployed to help slow down the dragster. Knowing that this deceleration is defined by the relationship a = - 0.022v^2, where v is the velocity in m/s, determine the total time from the beginning of the race until the car slows back down to 10 m/s and the total distance the car travels during this time. The total time from the beginning of the race until the car slows back down to 10 m/s is____________ s. (Round the final answer to one decimal place.) The total distance the car travels during this time is m. (Round the final answer to one decimal place.)
Physics
2 answers:
zloy xaker [14]2 years ago
8 0

Answer:

The total time is 9.2 s

Explanation:

For the first stage:

a = 50 - 10t

dv/dt = 50 - 10t

Integrating:

\int\limits^v_0 {} \, dv =\int\limits^t_0 {(50-10t)} \, dt \\v=50t-5t^{2}

v = 125 m/s

Replacing:

t² - 10t + 25 = 0

Solving, t = 5 s (time elapsed)

dx/dt = 50t - 5t²

Integrating:

\int\limits^x_0 {} \, dx =\int\limits^5_0 {50t} \, dt -\int\limits^5_0 {5t^{2} } \, dt \\x=\frac{50t^{2} }{2} -\frac{5t^{3} }{3} =\frac{1250}{3} =416.67m

For the second stage:

a=-0.022v^{2} \\\frac{dv}{dt} =-0.022^{2} \\\int\limits^a_b {\frac{dv}{v^{2} } } \,  =-0.022\int\limits^t_0 {} \, dt , where a=10, b=125\\\frac{1}{10} -\frac{1}{125} =0.022t\\t=4.2s

\frac{dv}{dx} *\frac{dx}{dt} =-0.022v^{2} \\v\frac{dv}{dx} =-0.022v^{2} \\\frac{dv}{dx} =-0.022v\\\int\limits^a_b {\frac{dv}{v} =-0.022} \,  \int\limits^x_0 {} \, dx ,where-a=10,b=125\\ln\frac{10}{125} =-0.022x

Resolving x = 114.8 m

The total time is 9.2 s

The total distance is 531.8 m

xz_007 [3.2K]2 years ago
5 0

Answer:

Mistake in question

The correct question

A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10t , where a and t are in m/s² and seconds, respectively. After reaching a speed of 125 m/s, a parachute is deployed to help slow down the dragster. Knowing that this deceleration is defined by the relationship a = - 0.02v², where v is the velocity in m/s, determine (a) the total time from the beginning of the race until the car slows back down to 10 m/s, (b) the total distance the car travels during this time.

Explanation:

Given the function

a = 50 —10t

The car started from rest u = 0

And it accelerates to a speed of 125m/s

Then, let find the time in this stage

Acceleration can be modeled by

a = dv/dt

Then, dv/dt = 50—10t

Using variable separation to solve the differentiation equation

dv = (50—10t)dt

Integrating both sides

∫ dv = ∫ (50—10t)dt

Note, v ranges from 0 to 125seconds, so we want to know the time when it accelerate to 125m/s. So t ranges from 0 to t'

∫ dv = ∫ (50—10t)dt

v = 50t —10t²/2. Equation 1

[v] 0<v<125 = 50t —10t²/2 0<t<t'

125—0 = 50t — 5t² 0<t<t'

125 = 50t' — 5t'²

Divide through by 5

25 = 10t' — t'²

t'² —10t' + 25 = 0

Solving the quadratic equation

t'² —5t' —5t' + 25 = 0

t'(t' —5) —5(t' + 5) = 0

(t' —5)(t' —5) = 0

Then, (t' —5) = 0 twice

Then, t' = 5 seconds twice

So, the car spent 5 seconds to get to 125m/s.

The second stage when the parachute was deployed

We want to the time parachute reduce the speed from 125m/s to 10m/s,

So the range of the velocity is 125m/s to 10m/s. And time ranges from 0 to t''

The function of deceleration is give as

a = - 0.02v²

We know that, a = dv/dt

Then, dv/dt = - 0.02v²

Using variable separation

(1/0.02v²) dv = - dt

(50/v²) dv = - dt

50v^-2 dv = - dt

Integrate Both sides

∫ 50v^-2 dv = -∫dt

(50v^-2+1) / (-2+1)= -t

50v^-1 / -1 = -t

- 50v^-1 = -t

- 50/v = - t

Divide both sides by -1

50/v = t. Equation 2

Then, v ranges from 125 to 10 and t ranges from 0 to t''

[ 50/10 - 50/125 ] = t''

5 - 0.4 = t''

t'' = 4.6 seconds

Then, the time taken to decelerate from 125s to 10s is 4.6 seconds.

So the total time is

t = t' + t''

t = 5 + 4.6

t = 9.6 seconds

b. Total distanctraveleded.

First case again,

We want to find the distance travelled from t=0 to t = 5seconds

a = 50—10t

We already got v, check equation 1

v = 50t —10t²/2 + C

v = 50t — 5t² + C

We add a constant because it is not a definite integral

Now, at t= 0 v=0

So, 0 = 0 - 0 + C

Then, C=0

So, v = 50t — 5t²

Also, we know that v=dx/dt

Therefore, dx/dt = 50t — 5t²

Using variable separation

dx = (50t —5t²)dt

Integrate both sides.

∫dx = ∫(50t —5t²)dt

x = 50t²/2 — 5 t³/3 from t=0 to t=5

x' = [25t² — 5t³/3 ]. 0<t<5

x' = 25×5² — 5×5³/3 —0

x' = 625 — 208.333

x' = 416.667m

Stage 2

The distance moved from

t=0 to t =4.6seconds

a = -0.002v²

We already derived v(t) from the function above, check equation 2

50/v = t + C.

When, t = 0 v = 125

50/125 = 0 + C

0.4 = C

Then, the function becomes

50/v = t + 0.4

50v^-1 = t + 0.4

Now, v= dx/dt

50(dx/dt)^-1 = t +0.4

50dt/dx = t + 0.4

Using variable separation

50/(t+0.4) dt = dx

Integrate both sides

∫50/(t+0.4) dt = ∫ dx

50 In(t+0.4) = x

t ranges from 0 to 4.6seconds

50In(4.6+0.4)—50In(4.6-0.4) = x''

x'' = 50In(5) —50In(4.2)

x'' = 8.72m

Then, total distance is

x = x' + x''

x = 416.67+8.72

x = 425.39m

The total distance travelled in both cases is 425.39m

You might be interested in
a deer with a mass of 176 kg is running head-on towards you with a velocity of 19 m/s. you are going north. find the magnitude a
Igoryamba

Momentum = Mass × Velocity

According to this formula,

Momentum of deer = 176 × 19 = 3344 kg•m/s.

Since you are heading north and the deer is running towards you, the direction of the deer' s momentum is north as well.

6 0
3 years ago
How does friction help soccer players
Lena [83]
When soccer players run they are using friction to propell themselves
7 0
3 years ago
Read 2 more answers
What other places in the world use hydro kinetic energy to generate electrical power?
Elina [12.6K]

Answer:

Alaska: Hydrokinetic Energy Campbell CR9000X used for in-stream hydrokinetic device evaluation. Marine hydrokinetic energy power generation is an emerging sector in the renewable energy portfolio. Hydrokinetic devices convert the energy of waves, tidal currents, ocean currents or river currents into electrical power.

3 0
3 years ago
What do you call the height of a wave?
Lina20 [59]

Answer:

amplitude is the answer

5 0
3 years ago
Read 2 more answers
A 7.00 kg bowling ball is held 2.00 m above the ground. Using g = 9.80 m/s2, how much energy does the bowling ball have due to i
deff fn [24]
mgh =7*2*9.8=137.2. It should be the answer.
4 0
3 years ago
Read 2 more answers
Other questions:
  • Rutherford's gold-foil experiment led him to conclude that
    11·1 answer
  • BWhat Do You Get When You Multiply An Object's mass times the acceleration?
    13·1 answer
  • What is the effective resistance of this dc circuit
    5·1 answer
  • HEY can anyone tell me the Atomic Mass of Helium and also what ever the number u get pls round it!!
    11·2 answers
  • A kangaroo jumps up with an initial velocity of
    12·1 answer
  • At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is
    8·1 answer
  • What lasts longer and why?: A solar eclipse or a lunar eclipse?
    11·1 answer
  • A skateboarder travels on a horizontal surface with an initial velocity of 3.6 m/s toward the south and a constant acceleration
    5·1 answer
  • A volume of 25 milliliters is the same as a volume of ____________________ cubic centimeters.
    5·1 answer
  • A small box is held in place against a rough vertical wall by someone pushing on it with a force directed upward at 27 ∘ above t
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!