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anastassius [24]
3 years ago
5

a person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500m/s^2. what is the velocity of the stroll

er after it has traveled 4.75m? 
Physics
2 answers:
Zigmanuir [339]3 years ago
6 0

Hello!

A person pushing a stroller start from rest uniformly accelerating at a rate of 0.500 m/s². What is the velocity of the stroller after it traveled 4.75 m ?

We have the following data:

a (acceleration) = 0.500 m/s²

Vf (final velocity) = ? (in m/s)

Vi (initial velocity) = 0 m/s

Δx (displacement) = 4.75 m

Solving:

Let's apply the Torricelli Equation, to find the velocity of the stroller, let's see:

V_f^2 = V_i^2 + 2*a*\Delta{x}

V_f^2 = 0^2 + 2*0.500*4.75

V_f^2 = 0 + 4.75

V_f^2 = 4.75

V_f = \sqrt{4.75}

\boxed{\boxed{V_f \approx 2.18\:m/s}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}

Answer:

The velocity of the stroller is approximately 2.18 m/s

________________________________

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}

grandymaker [24]3 years ago
5 0
You're going to use the constant acceleration motion equation for velocity and displacement:
(V)final²=(V)initial²+2a(Δx)

Given:
a=0.500m/s²
Δx=4.75m
(V)intial=0m
(V)final= UNKNOWN

(V)final= 2.179m/s
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We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

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Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2}  } \\

F_{m} =G*\frac{m_{m}*m_{a}  }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]

When we match these equations the masses cancel out as the universal gravitational constant

G*\frac{m_{e} *m_{a} }{r_{e}^{2}  } = G*\frac{m_{m} *m_{a} }{r_{m}^{2}  }\\\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2}  }

To solve this equation we have to replace the first equation of related with the distances.

\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m}  )^{2}  } = \frac{7.36*10^{22}  }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17}  \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c }  }{2*a}\\  where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) }  }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

<u />

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2}  } \\a=3.33*10^{19} [m/s^2]

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Substitute these value into equation 1

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