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natka813 [3]
3 years ago
9

A car braked with a constant deceleration of 16ft/s2, producing skid marks measuring 200 feet before coming to a stop. How fast

was the car traveling (in ft/s) when the brakes were first applied
Mathematics
1 answer:
Llana [10]3 years ago
7 0

Answer:the car was traveling at a speed of 80 ft/s when the brakes were first applied.

Step-by-step explanation:

The car braked with a constant deceleration of 16ft/s^2. This is a negative acceleration. Therefore,

a = - 16ft/s^2

While decelerating, the car produced skid marks measuring 200 feet before coming to a stop.

This means that it travelled a distance,

s = 200 feet

We want to determine how fast the car was traveling (in ft/s) when the brakes were first applied. This is the car's initial velocity, u.

Since the car came to a stop, its final velocity, v = 0

Applying Newton's equation of motion,

v^2 = u^2 + 2as

0 = u^2 - 2 × 16 × 200

u^2 = 6400

u = √6400

u = 80 ft/s

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Hey there!

ANSWER: \frac{23}{20}

EXPLANATION:

To find the answer to your question, you will need to add.

\frac{3}{4}+\frac{2}{5}

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5*4=20

So the denominator will be 20.

Now move on to the numerator. Multiply 5 by 3.

5*3=15

Now multiply 4 bu 2.

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If you want to get the numerator, add what you got after multiply.

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\frac{23}{20}

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victus00 [196]

Answer:

x = 0 , 3

Step-by-step explanation:

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OleMash [197]

Answer:

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Step-by-step explanation:

Given

See attachment

Required

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From the attachment, we have:

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Take L,C,M

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andrey2020 [161]
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Burka [1]

Answer:

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Step-by-step explanation :

Answer:

The red segment is from 1/4 to 3/4.

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