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3 years ago

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A horse and rider are racing to the right with a speed of 21\,\dfrac{\text m}{\text s}21 s m 21, start fraction, start text, m

, end text, divided by, start text, s, end text, end fraction when they pass the finish line and begin slowing down. The horse slows for 52\,\text m52m52, start text, m, end text with constant acceleration before it stops. What was the acceleration of the horse as it came to a stop? Answer using a coordinate system where rightward is positive. Round the answer to two significant digits.

2 answers:

abruzzese [7]3 years ago

8 0

Answer:

The acceleration of the horse as it came to a stop is $4.24\ m/s^2$ .

Explanation:

Given that,

Initial speed of the horse and rider, u = 21 m/s

It finally comes to rest, v = 0

The horse slows down for 52 m with constant acceleration before it stops.

We need to find the acceleration of the horse as it came to a stop. Let a is the acceleration. We can find it using third equation of motion as :

We need to find the acceleration of the horse as it came to a stop. Let a is the acceleration. We can find it using third equation of motion as :

$v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{-u^2}{2d}\\\\a=\dfrac{-(21)^2}{2\times 52}\\\\a=-4.24\ m/s^2$

So, the acceleration of the horse as it came to a stop is $4.24\ m/s^2$ . Negative sign shows deceleration.

Marrrta [24]3 years ago

7 0

Answer:

a=-4.2 m/s²

Explanation:

The horse riding so inital velocity is given finally the rider stops so the final velocity is zero.

initial velocity =Vi= 21 m/s

final velocity =Vf= 0 m/s

distance covered = S=52 m

By using 2nd equation of motion we can find the acceleration

2aS=Vf² -Vi²

a=(-441)/104

a=-4.2 m/s²

So the accceleration is 4.2 m/s².

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Answer:

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Explanation:

This is an exercise that we can solve with kinematics equations.

Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.

x axis

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initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration

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ay = (v_{oy} -v_{y}) / t

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Answer:

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