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Katyanochek1 [597]
2 years ago
8

A rocket sled accelerates from rest for a distance of 645 m at 16.0 m/s2. A parachute is then used to slow it down to a stop. If

the parachute gives the sled an acceleration of -18.2 m/s2 and there is 500.0 m of sled track remaining after the shoot opens, will the sled stop before running off the track? Show why or why not?
Physics
1 answer:
inessss [21]2 years ago
3 0

Answer:

the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

Explanation:

This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period

          v² = v₀² + 2 a₁ x

indicate that the initial velocity is zero

          v² = 2 a₁ x

let's calculate

          v = \sqrt {2 \ 15.0 \ 645}

          v = 143.666 m / s

now for the second interval let's find the distance it takes to stop

          v₂² = v² - 2 a₂ x₂

in this part the final velocity is zero (v₂ = 0)

         0 = v² - 2 a₂ x₂

         x₂ = v² / 2a₂

let's calculate

         x₂ = \frac{ 143.666^2 }{2 \  18.2}

         x₂ = 573 m

as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

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Answer:

F=-18412.9N, where the minus indicates the direction is opposite to that of the throw.

Explanation:

a)

Since MKS stands for meter-kilogram-second and we know that:

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So we have that:

90 mph=90 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=40m/s

110 mph=110 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=48.89m/s

145 g=145 g(\frac{1kg}{1000g})=0.145kg

b)

Newton's 2nd Law tells us that F=ma, and the definition of acceleration is a=\frac{\Delta v}{\Delta t}, so we have:

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Taking the throw direction as the positive one, for our values we have:

F=m\frac{v_f-v_i}{t}=(0.145kg)\frac{(-48.89m/s)-(+40m/s)}{0.0007s}=-18412.9N

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