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Studentka2010 [4]
4 years ago
13

A speeder is driving down the road at a constant 15 m/s, passes a police officer parked on the roadside. The officer pauses 9 se

conds, then pursues the speeder, accelerating at a constant 5 m/s^2.
How much time does it take the police officer to catch the speeder?
How far did the police officer drive before the speeder was caught?
Physics
1 answer:
Amanda [17]4 years ago
6 0

Answer:

3+3\sqrt{7} seconds

299.0588 meters

Explanation:

Given

  • speeder has a constant speed,x=15
  • officer starts 9 seconds after speeder crosses him and accelerates at 5 from rest

Let assume S as the distance covered by police before he catches him

let T be the time taken by him to do so

(distance=\text{initial velocity}\times time+\frac{1}{2} \times acceleration\times time)

Therefore S=\frac{1}{2} aT^{2} =\frac{5}{2} T^{2}(since initial velocity=0)

This same distance is covered by the speeder in time T+9 as officer starts after pausing 9 seconds

Therefore S= (T+9)\times 15=15T+ 135

equating both the equations

\frac{5}{2}}T^{2}=15T+ 135\\5T^{2}=30T+270\\T^{2}=6T+54\\T^{2}-6T-54=0

Solving the quadratic we get

T=3+3\sqrt{7} or T=3-3\sqrt{7}(not possible as T cannot be less than 0)

So it takes 3+3\sqrt{7} seconds for the officer to catch the speeder

⇒Distance covered =\frac{5}{2} T^{2}=299.0588 m

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Answer:

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b. T_f=86.4^{\circ}C

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Explanation:

Given:

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  • quantity of water suggested for the given cheese, m_w=3\ quarts\times 946.35=2839.05\ g
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a.

  • we have specific heat of water, c_w=1\ cal.g^{-1}.^{\circ}C^{-1}

so now, we use the heat equation to find the amount of heat exchange during the process:

m_c\times c_c\times \Delta T=m_w\times c_w\times \Delta T

255\times 0.4\times (T_f-(-40))=2839.05\times 1\times (100-T_f)

T_f=95.14^{\circ}C

b.

when using 1 quart of water:

255\times 0.4\times (T_f-(-40))=946.35\times 1\times (100-T_f)

T_f=86.4^{\circ}C

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Explanation:

Given that,

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\theta=2\pi\times46

\theta=289.02\ rad

We need to calculate the angular acceleration

Using equation of motion

\omega_{f}^2=\omega_{i}^2+2\alpha\theta

\alpha=\dfrac{\omega_{f}^2-\omega_{i}^2}{2\theta}

\alpha=\dfrac{0-(366)^2}{2\times289.02}

\alpha=-231.74\ rad/s^2

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