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zimovet [89]
3 years ago
7

Please please please please please help me!!!!!​

Mathematics
1 answer:
Bond [772]3 years ago
6 0

Answer:

c

Step-by-step explanation: just trust me.

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Step-by-step explanation:

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Answer:

Refer to your notes from module 6 on this.

Step-by-step explanation:

Complete this test question just like you did the practice question from the module. Remember your test is an open note test, but it is not a copy from the internet test. You can do it!!!

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A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2.
IRISSAK [1]

Answer:

80 feet

Step-by-step explanation:

Given:

Initial speed of the car (v_0) = 40 ft/sec

Deceleration of the car (\frac{dv}{dt}) = -10 ft/sec²

Final speed of the car (v_x) = 0 ft/sec

Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.

Now, deceleration means rate of decrease of velocity.

So, \frac{dv}{dt}=-10\ ft/sec^2

Negative sign means the velocity is decreasing with time.

Now, \frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt}) using chain rule of differentiation. Therefore,

\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx

Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,

\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft

Therefore, the car travels a distance of 80 feet before stopping.

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3 years ago
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