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2 years ago

7

in what distance can a 1500kg automobile be stopped if the brake is applied when the speed is 20m/s and the coefficient of slidi

ng friction is 0.7 between the tyres and the ground?please I need a full step by step explanation

1 answer:

natali 33 [55]2 years ago

8 0

1) The braking force is provided by the frictional force, which is given by:

$F_f=\mu m g$

where

$\mu=0.7$ is the coefficient of friction

m=1500 kg is the mass of the car

$g=9.81 m/s^2$ is the gravitational acceleration

Substituting numbers into the equation, we find

$F_f= (0.7)(1500 kg)(9.81 m/s^2)=10301 N$

2) The work done by the frictional force to stop the car is equal to the product between the force and the distance d:

$W=-F_fd$ (1)

where we put a negative sign because the force is in the opposite direction of the motion of the car.

3) For the work-energy theorem, the work done by the frictional force is equal to the variation of kinetic energy of the car:

$\Delta K=K_f -K_i =W$ (2)

The final kinetic energy is zero, so the variation of kinetic energy is just equal to the initial kinetic energy of the car:

$\Delta K=-K_i=-\frac{1}{2}mv^2=-\frac{1}{2}(1500 kg)(20 m/s)^2=-300000 J$

4) By equalizing eq. (1) and (2), we find the distance, d:

$-K_i = -Fd$

$d=\frac{K_i}{F}=\frac{300000 J}{10301 N}=29.1 m$

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Answer:

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75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

So

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

2)

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

So

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

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