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Sonja [21]
2 years ago
15

The driver of a 1000 kg car traveling at a speed of 16.7 m/s applies the brakes. If the brakes provide a force of - 8000 N to st

op the car, how far did the car travel before it stopped
Physics
1 answer:
RSB [31]2 years ago
6 0

Answer:

Given:

m=1000kg

u= 16.7m/s

v=0m/s

F=8000N

Required:

s=?

Solution:

F=m × a

8000N=1000kg × a

a=8m/s^2

Since it decelerate a= -8m/s^2

v^2 = u^2 + 2as

s=v^2 - u^2 / 2a

s= 0 - (16.7m/s)^2 / 2 × -8m/s^2

s= -278.89/-16

s= 17.43m

The car travels approximately 17.43m before it stops

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The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

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We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

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u_y is the initial vertical velocity of the ball, which is given by

u_y = u sin \theta

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Substituting everything into the equation we get:

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Solving the equation for t, we find the time of flight of the ball:

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We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

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Therefore, the horizontal distance covered in a time t is

d=v_x t

And substituting t = 3.59 s, we find

d=(19.6)(3.59)=70.4 m

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Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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