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Tamiku [17]
3 years ago
6

A shopping cart rolls from x=19.9m to x=5.40m with an average velocity of -0.418m/s. How much time did it take? (unit=s)

Physics
1 answer:
andrew11 [14]3 years ago
8 0

Given

initial position = Xi= 19.9m

Final position Xf = 5.4m

Average velocity= Va = -0.418m/s

it shows displacement is reverse.

To find  t=?

As   Va = (Xf- Xi) / t

t = (Xf-Xi) / ( Va)

t = ( 5.4-19.9) / (-0.418)

t = (-14.5 ) / (-0.418)   (-ve sign cancel out at numerator and denominator)

t =34.69 s

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yuradex [85]

Answer:

9) a = 25 [m/s^2], t = 4 [s]

10) a = 0.0875 [m/s^2], t = 34.3 [s]

11) t = 32 [s]

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9)

a)

v_{f}^{2} = v_{i}^{2}-(2*a*x)\\

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = acceleration [m/s^2]

x = distance = 200 [m]

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0 = (100)^2 - (2*a*200)

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b)

Now using the following equation:

v_{f} =v_{i} - (a*t)

0 = 100 - (25*t)

t = 4 [s]

10)

a)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

Note:  The positive sign of the equation means that the car increases his speed.

5^2 = 2^2 + 2*a*(125 - 5)

25 - 4 = 2*a* (120)

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b)

Now using the following equation:

v_{f}= v_{i}+a*t\\

5 = 2 + 0.0875*t

3 = 0.0875*t

t = 34.3 [s]

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To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

10^2 = 2^2 + 2*a*(200 - 10)

100 - 4 = 2*a* (190)

a = 0.25 [m/s^2]

Now using the following equation:

v_{f}= v_{i}+a*t\\

10 = 2 + 0.25*t

8 = 0.25*t

t = 32 [s]

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