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kirill115 [55]
3 years ago
9

A sprinter explodes out of the starting block with an acceleration of +4.2 m/s2, which she sustains for 0.9 s. Then, her acceler

ation drops to zero for the rest of the race. (a) What is her velocity at t = 0.9 s. m/s (b) What is her velocity at the end of the race? m/s
Physics
1 answer:
ad-work [718]3 years ago
6 0

Answer:

a) 3.78 m/s

b) 3.78 m/s

Explanation:

a )From the equations of kinematics   we know that

     Vf - Vi = at

since Initial speed Vi = 0

acceleration = 4.2 m/s2

so we have

     Vf = a t

           = (4.2) (0.9)

         = 3.78 m/s

velocity at t = 0.9 s. m/s is 3.78 m/s

b) If the sprinter maintains constant velocity then acceleration becomes zero.

   So velocity is 3.78 m/s

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Answer:

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Explanation:

Given:

The mass of the thrower, m_{t} = 70~Kg.

The mass of the catcher, m_{c} = 55~Kg.

The mass of the ball, m_{b} = 0.0480~Kg.

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Consider that the final velocity of the thrower is v_{ft}. From the conservation of momentum,

&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s

Consider that the final velocity of the catcher is v_{fc}. From the conservation of momentum,

&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s

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