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AleksandrR [38]
4 years ago
11

A turtle and a rabbit are in a 150 meter race. The rabbit decides to give the turtle a 1 minute head start. The turtle moves at

a constant speed of 0.500 m/s through the whole race (in fact the turtle even starts at a velocity of 0.500 m/s as while he was still approaching the starting line he was allowed to continue to keep going without stopping). The rabbit starts the race from rest and accelerates at a rate or 1.50 m/s2 until she reaches her top speed of 10 m/s. She then finishes the race running at a constant speed of 10 m/s. a) What is the turtle’s position when the rabbit starts to run (1 minute into the race)? b) How long does it take the turtle to finish the race? c) How long does it take the rabbit to reach max speed? d) What is the rabbit’s position when she reaches max speed? e) How long does it take the rabbit to finish the race? f) Who won?
Physics
1 answer:
yan [13]4 years ago
7 0

Answer:

a) s_{T} = 30\,m, b) t = 5\,min, c) \Delta t = 6.667\,s, d) \Delta s_{R} = 33.333\,m, e) t' = 11.667\,s, f) The rabbit won the race.

Explanation:

a) As turtle moves at constant speed, its position is determined by the following formula:

s_{T} = v_{T}\cdot t

Where:

t - Time, measured in seconds.

v_{T} - Velocity of the turtle, measured in meters per second.

s_{T} - Position of the turtle, measured in meters.

Then, the position of the turtle when the rabbit starts to run is:

s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)

s_{T} = 30\,m

The position of the turtle when the rabbit starts to run is 30 meters.

b) The time needed for the turtle to finish the race is:

t = \frac{s_{T}}{v_{T}}

t = \frac{150\,m}{0.5\,\frac{m}{s} }

t = 300\,s

t = 5\,min

The time needed for the turtle to finish the race is 5 minutes.

c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:

v_{R} = v_{o,R} + a_{R}\cdot \Delta t

Where:

v_{R} - Final velocity of the rabbit, measured in meters per second.

v_{o,R} - Initial velocity of the rabbit, measured in meters per second.

a_{R} - Acceleration of the rabbit, measured in \frac{m}{s^{2}}.

\Delta t - Running time, measured in second.

\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}

\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }

\Delta t = 6.667\,s

The time taken by the rabbit to reach maximum speed is 6.667 s.

d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:

v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}

\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}

\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}

Where \Delta s_{R} is the travelled distance of the rabbit from rest to maximum speed.

\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}

\Delta s_{R} = 33.333\,m

The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.

e) The time required for the rabbit to finish the race can be determined by the following expression:

t' = \frac{\Delta s_{R}}{v_{R}}

t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }

t' = 11.667\,s

The time required for the rabbit from rest to maximum speed is 11.667 seconds.

f) The animal with the lowest time wins the race. Now, each running time is determined:

Turtle:

t_{T} = 300\,s

Rabbit:

t_{R} = 60\,s + 6.667\,s + 11.667\,s

t_{R} = 78.334\,s

The rabbit won the race as t_{R} < t_{T}.

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In an elastic collision, the total kinetic energy is conserved, while in an inelastic collision, it is not

Explanation:

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A thin, metallic spherical shell of radius 0.227 m has a total charge of 6.03 × 10 − 6 C placed on it.
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Answer:

Explanation:

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radius r=0.227 m

Charge on surface Q=6.03\times 10^{-6} C

Point Charge inside sphere q=1.15\times 10^{-6} C

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Treating Surface charge as Point charge and applying Gauss law

E_{total}A=\frac{q_{enclosed}}{\epsilon _0}

where A=surface area up to distance r

E_{total}=\frac{Q+q}{4\pi r^2}

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Answer:

W=1705.2 J

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Now by putting the values

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W= 60 x 9.8 x 2.9 J

W=1705.2 J

Therefore the answer will be 1705.2 J.

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