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3 years ago

You’re speeding at 85 km/hr when you notice that you’re only 10 m behind the car in front of you,

1 answer:

7 0

Your position in meters will, measured relative to the starting point of the car behind you, be
x1(t) = 10 + 23.61 t - 1/2 4.2 t^2
his position will be
x2(t) = 16.67 t
Hence at any time the separation s(t) will be
s(t) = x1(t) - x2(t) = 10 + 6.94 t -2.1 t^2
Now I assume you mean that you will decelerate UNTIl you are driving at the legal speed limit (60 km/h). That will take you:
16.67 m/s = 23.61m/s - 4.2 m/s^2 * t
t = 1.65 seconds
What is the separation at that time? If it is still greater than zero, there will be no collision:
s(1.65) = 10 + 6.94 *1.65 -2.1 (1.65)^2 = 15.73 meters.
Hence you will NOT collide. The 1.65 s you calculated was the time needed to brake to the speed of 60 km/h.

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4 years ago

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Answer:

Part a)

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

$\tau = I\alpha$

here we will have

$\tau = (m_1g - m_2g)(\frac{L}{2})$

now we will have inertia of two masses given as

$I = (m_1 + m_2)(\frac{L}{2})^2$

now we have

$I = (m_1 + m_2)\frac{L^2}{4}$

now the angular acceleration is given as

$\alpha = \frac{\tau}{I}$

so we have

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

Now if the rod is not massles then we will have total inertia given as

$I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}$

so we will have

$I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}$

now the acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

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4 years ago

Un vehicle surt, en un instant, d'un punt A cap a un altre punt B amb una velocitat constant de 36 km/h. Després d'un minut(60 s

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3 years ago

If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the shortest distance in which you can st

liberstina [14]

Answer:

The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

Explanation:

Given;

coefficient of kinetic friction, μ = 0.84

speed of the automobile, u = 29.0 m/s

To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration

x is the shortest distance

First we determine a;

From Newton's second law of motion

∑F = ma

F is the kinetic friction that opposes the motion of the car

-Fk = ma

but, -Fk = -μN

-μN = ma

-μmg = ma

-μg = a

- 0.8 x 9.8 = a

-7.84 m/s² = a

Now, substitute in the value of a in the equation above

v² = u² + 2ax

when the automobile stops, the final velocity, v = 0

0 = 29² + 2(-7.84)x

0 = 841 - 15.68x

15.68x = 841

x = 841 / 15.68

x = 53.64 m

Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

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3 years ago