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USPshnik [31]
3 years ago
8

HURRY!!!!

Physics
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

Ted is correct

Explanation:

The equation for gravitational potential energy is PE = m·g·h

The equation for gravitational kinetic energy is KE = 1/2·m·v²

Where:

m = Mass of the object (The racing car)

g = Acceleration due to gravity

h = The height to which the object is raised

v = Velocity of motion of the object

From the principle of conservation of energy, energy can neither be created nor destroyed but changes from one form to another, we have;

Potential energy gained from location at height h = Kinetic energy gained as the object moves down the level ground

m·g·h = 1/2·m·v² canceling like terms gives

g·h = 1/2·v²

v = (√2·g·h)

If the speed is doubled, we have

2·v = 2× (√2·g·h) =  (√2·g·4·h)

Therefore, if 2·v = v₂ then v₂ =  (√2·g·4·h)

Since g, the acceleration due to gravity, is constant, it means that the initial  height must be multiplied or increased 4 times to get the new height, that is we have;

v₂ =  (√2·g·4·h) = (√2·g·h₂)

Where:

4·h = h₂

Which gives;

v₂² = 2·g·h₂

1/2·v₂² = g·h₂

1/2·m·v₂² = m·g·h₂ Just like in the first relation

Therefore, Ted is correct s they need to go up four times the initial height to double the speed.

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\blue{\bold{\underline{\underline{Answer:}}}}

\green{\tt{\implies{Time=10\:sec\:and\:30\:sec}}} \\\\

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}} \\ \\

\green{\underline{\bold{Given :}}}  \\ \\   \:\:\:\: \bullet\:\:\tt\red{ Velocity \: of \: boy = 50 \: m/s} \\  \\   \:\:\:\: \bullet\:\: \tt\orange{Velocity \: of \: girl = 30 \: m/s }\\  \\  \:\:\:\: \bullet\:\:\tt\green{ Acceleration \: of \: boy =  {1 \: m/s}^{2}} \\  \\  \:\:\:\:\bullet\:\: \tt\blue{Acceleration \: of \:girl=  {2\: m/s}^{2} }\\  \\ \:\:\:\: \bullet \:\:\tt\purple{Sepration \: between \: them = 150 \: m }\\ \\  \\ \red{\underline{\bold{To \: Find :}}} \\ \\  \:\:\:\: \bullet\:\: \tt\blue{Time \: taken \: to \: catch \: the \: girl  }\\

<u>According to given question</u> :

\\ \green{\star} \:  \text{Using \: relative \: motion \: method} \\  \\  \green{ \circ} \:  \tt Net \: velocity = 50 - 30 = 20 \: m/s \\  \\ \green{ \circ} \:  \tt Net \: acceleration = 1 - 2 = - 1\: m /{s}^{2}  \\\\  \\  \star\:\bold\red{\underline{\:As \: we \: know \: that\:}} \\\\  \tt\purple{:  \implies s = ut +  \frac{1}{2}  {at}^{2}}  \\  \\ \tt\green{:  \implies 150 = 20 \times t +  \frac{1}{2}  \times  -1 \times  {t}^{2}}  \\  \\ \tt\purple{:  \implies 300 = 40t -  {t}^{2}}  \\  \\ \tt\green{:  \implies  {t}^{2}  - 40t  + 300 = 0} \\  \\ \tt\purple{:  \implies t =  \frac{  - ( - 40) \pm\sqrt{ { (- 40)}^{2}  - 4 \times 1 \times 300} }{2 \times 1}  }\\  \\ \tt\green{:  \implies t =  \frac{40 \pm \sqrt{1600 - 1200} }{2}  }\\  \\ \tt\purple{:  \implies t =  \frac{40 \pm 20}{2}  }\\  \\  \green{\tt:  \implies t = 10 \: sec \: and \: 30 \: sec}

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