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Lyrx [107]
3 years ago
11

Willie, in a 100.0 m race, initially accelerates uniformly from rest at 2.00 m/s2 until reaching his top speed of 12.0 m/s. He m

aintained this speed until he is 16.0 m from the finish line but then fades and decelerates uniformly, crossing the line with a speed of only 8.00 m/s. What is Willie’s total time for the race? Please show your steps :)
Physics
1 answer:
Oduvanchick [21]3 years ago
6 0

Answer:

The total time for the race is 11.6 seconds

Explanation:

The parameters given are;

Total distance ran by Willie = 100.0 m

Initial acceleration = 2.00m/s²

Top speed reached with initial acceleration = 12.0 m/s

Point where Willie start to fade and decelerate = 16.0 m from the finish line

Speed with which Willie crosses the finish line = 8.00 m/s

The time and distance covered with the initial acceleration are found using the following equations of motion;

v = u₀ + a·t

v² = u₀² + 2·a·s

Where:

v = Final velocity reached with the initial acceleration = 12.0 m/s

u₀ = Initial velocity at the start of the race = 0 m/s

t = Time during acceleration

a = Initial acceleration = 2.00 m/s²

s = Distance covered during the period of initial acceleration

From, v = u₀ + a·t, we have;

12 = 0 + 2×t

t = 12/2 = 6 seconds

From, v² = u₀² + 2·a·s, we have;

12² = 0² + 2×2×s

144 = 4×s

s = 144/4 =36 meters

Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;

Distance covered at top speed = 100 - 36 - 16 = 48 meters

Time, t_t of running at top speed = Distance/velocity = 48/12 = 4 seconds

The deceleration from top speed to crossing the line is found as follows;

v₁² = u₁² + 2·a₁·s₁

Where:

u₁ = v = 12 m/s

v₁ = The speed with which Willie crosses the line = 8.00 m/s

s₁ = Distance covered during decelerating = 16.0 m

a₁ = Deceleration

From which we have;

8² = 12² + 2 × a × 16

64 = 144 + 32·a

64 - 144 = 32·a

32·a = -80

a = -80/32 = -2.5 m/s²

From, v₁ = u₁ + a₁·t₁

Where:

t₁ = Time of deceleration

We have;

8 = 12 + (-2.5)·t₁

t₁ = (8 - 12)/(-2.5) = 1.6 seconds

The total time = t + t_t + t₁ =6 + 4 + 1.6 = 11.6 seconds.

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\frac{1}{10000} + \frac{1}{2000} + \frac{1}{1000} \\ \\ = \frac{1}{1000} ( \frac{1}{10} + \frac{1}{2} + \frac{1}{1} ) \\ \\ = \frac{1}{1000} ( \frac{31}{20}) \\ \\ = \frac{31}{20000}

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3 years ago
A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn
77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}

u = -21.09 cm

The magnification of the mirror is given by :

m=\dfrac{-v}{u}

m=\dfrac{-(-37.5)}{(-21.09)}

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

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52,360,000km

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To solve this problem you use a conversion factor.

By taking into account that 1UA = 1.496*10^{8}km you obtain:

0.35UA*\frac{1.496*10^{8}km}{1UA}=52,360,000km

hence, 0.35UA is about 52,360,000km. This is the least distance between Mars and Earth

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denis-greek [22]

Answer:

A 300-kg crate is placed on a adjustable incline plane. as one end of the incline is raised, the crate begins to move downward. if the crate slides down the plane with the acceleration of .70 m/s2 , when the incline is 25 degrees, what is the coefficient of kinetic friction between ramp and crate?

Explanation:

6 0
3 years ago
A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t
Irina-Kira [14]

Answer:

part (a) \alpha\ =\ -2.5\ rad/s^2

part (b) N = 79.61 rev

part (c) \tau\ =\ 23.54\ Nm

Explanation:

Given,

  • Initial speed of the wheel = w_o\ =\ 50.0\ rad/s
  • total time taken = t = 20.0 sec

part (a)

Let \alpha be the angular acceleration of the wheel.

Wheel is finally at the rest. Hence the final angular speed of the wheel is 0.

\therefore w_f\ =\ w_0\ +\ \alpha t\\\Rightarrow \alpha\ =\ -\dfrac{w_0}{t}\\\Rightarrow \alpha\ =\ -\dfrac{50}{20}\\\Rightarrow \alpha\ =\ -2.5\ rad/s^2

part (b)

Let \theta be the total angular displacement of the wheel from initial position till the rest.

\therefore \theta\ =\ w_0t\ +\ \dfrac{1}{2}\alphat^2\\\Rightarrow \theta\ =\ 50\times 20\ -\ 0.5\times 2.5\times 20^2\\\Rightarrow \theta\ =\ 500\ rad

We know,  1 revolution = 2\pi rad

Let N be the number of revolution covered by the wheel.

\therefore N\ =\ \dfrac{\theta}{2\pi}\\\Rightarrow N\ =\ \dfrac{500}{2\times 3.14}\\\Rightarrow N\ =\ 79.61\ rev

Hence the 79.61 revolution is covered by the wheel in the 20 sec.

part (c)

Given,

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Now the center of mass of the pole = d\ =\ \dfra{L}{2}\ =\ \dfrac{2.5}{2}\ =\ 1.25\ m

Weight component of the pole perpendicular to the center of mass = F\ =\ mgcos\theta

\therefore \tau\ =\ F\times d\\\Rightarrow \tau\ =\ 4\times 9.81\times cos60^o\times 1.25\\\Rightarrow \tau\ =\ 23.54\ Nm

3 0
3 years ago
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