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A 6.0-kilogram cart initially traveling at 4.0 meters per second east accelerates uniformly at 0.50 meter per second squared east for 3.0 seconds. What is the speed of the cart at the end of this 3.0 second interval? A. 1.5 m/s B. 5.5 m/s G. 3.0 m/s D. 7.0 m/s​

Respuesta :

Answer: 5.5m/s

Explanation:

vf=vi+at

vf= 4.0m/s + (0.50m/s^2)(3.0s)

Lanuel

The speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.

Given the following data:

  • Initial velocity = 4 m/s
  • Mass of cart = 6 Kg
  • Acceleration = 0.5 [tex]m/s^2[/tex]
  • Time = 3 seconds

To find the speed of the cart at the end of this 3.0 second interval, we would use the first equation of motion;

[tex]V = U + at[/tex]

Where:

  • U is the initial velocity.
  • V is the final velocity.
  • a is the acceleration.
  • t is the time measured in seconds.

Substituting the given parameters into the formula, we have;

[tex]V = 4 + 0.5(3)\\\\V = 4 + 1.5[/tex]

Final velocity, V = 5.5 m/s.

Therefore, the speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.

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