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Mama L [17]
3 years ago
12

The driver of a car slams on the brakes, causing the car to slow down at a rate of 17ft/s2 as the car skids 285ft to a stop.

Physics
1 answer:
ozzi3 years ago
8 0

1) 5.79 s

2) 98.4 ft/s

Explanation:

1)

The motion of the car is a uniformly accelerated motion (it means it travels with constant acceleration), so we can find the time it takes for the car to stop by using the following suvat equation:

s=vt-\frac{1}{2}at^2

where

s is the distance travelled

v is the final velocity

t is the time

a is the acceleration of the car

In this problem we have:

s = 285 ft is the distance travelled

a=-17 ft/s^2 is the acceleration of the car (negative since the car is slowing down)

v = 0 ft/s is the final velocity of the car, since it comes to a stop

Solving for t, we find:

t=\sqrt{\frac{-2s}{a}}=\sqrt{\frac{-2(285)}{-17}}=5.79 s

2)

The initial speed of the car can be found by using another suvat equation, namely:

v=u+at

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

In this problem, we have:

v = 0 is the final speed of the car

a=-17 ft/s^2 is the acceleration of the car (negative since the car is slowing down)

t = 5.79 s is the total time of motion (found in part 1)

Therefore, the initial speed of the car is:

u=v-at=0-(-17)(5.79)=98.4 ft/s

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Artyom0805 [142]

2.50 miles is equal to 4026 m.

<u>Explanation:</u>

As it is known that 1000 m =1 km and 1 km = 0.621 miles. So first we have to convert miles to km and then to metre as follows.

As 1 km = 0.621 miles, then

             \text { 1 miles }=\frac{1}{0.621} \mathrm{km}

So, 2.50 miles will be equal to

            2.50 \text { miles }=\frac{2.50}{0.621} \mathrm{km}=4.026 \mathrm{km}

Then, in order to get the answer in meters, we have to convert this km to meter by the conversion of 1000 m =1 km.  So,

           1 \mathrm{km}=1000 \mathrm{m}

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          4.026 \mathrm{km}=4026 \mathrm{m}

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3 years ago
A squirrel jumps into the air with a velocity of 4 m/s at an angel of 50 degrees. What is the maximum height reached by the squi
Debora [2.8K]

Answer:

Explanation:

Assuming the squirrel is jumping off the ground, here's what we know but don't really know...

v₀ = 4.0 at 50.0°

So that's not really the velocity we are looking for. We are dealing with a max height problem, which is a y-dimension thing. Therefore, we need the squirrel's upward velocity, which is NOT 4.0 m/s. We find it in the following way:

v_{0y}=4.0sin(50.0) which gives us that the upward velocity is

v₀ = 3.1 m/s

Moving on here's what we also know:

a = -9.8 m/s/s and

v = 0

Remember that at the very top of the parabolic path, the final velocity is 0. In order to find the max height of the squirrel, we need to know how long it took him to get there. We are using 2 of our 3 one-dimensional equations in this problem. To find time:

v = v₀ + at and filling in:

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-3.1 = -9.8t so

t = .32 seconds.

Now that we know how long it took him to get to the max height, we use that in our next one-dimensional equation:

Δx = v_0t+\frac{1}{2}at^2 and filling in:

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Δx = .99 - .50 so

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3 years ago
THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!!
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Answer:

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I most sure about c but could be wrong
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kobusy [5.1K]

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Explanation:

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