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if a Firebird travels at a velocity of 0 to 60 mph in four seconds traveling east what was the acceleration of the Firebird

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VectorFundament120

Answer:

6.7 m/s^2

Explanation:

The formula of acceleration is:

[tex]\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{v_2 - v_1}{t_2-t_1}}[/tex]

where [tex]\displaystyle{\vec{a}}[/tex] is acceleration, [tex]\displaystyle{\vec{v}}[/tex] is velocity and [tex]\displaystyle{t}[/tex] is time. [tex]\displaystyle{v_2}[/tex] means final velocity. [tex]\displaystyle{v_1}[/tex] means initial velocity, [tex]\displaystyle{t_2}[/tex] means final time and [tex]\displaystyle{t_1}[/tex] means initial time.

We are given that the Firebird travels at velocity of 0 to 60 mph in four seconds. Therefore:

  • Our initial velocity starts at 0 mph.
  • Our final velocity is at 60 mph.
  • Our initial time is 0 second.
  • Our final time is 4 seconds.

Since it travels to the east then our vector will be positive. However, acceleration has to be in m/s^2 unit (Sl unit) so we'll have to convert from mph (miles per hours) to m/s (meters per second) first.

We know that:

  • A mile equals to 1609.344 meters.
  • An hour equals to 60 minutes which a minute equals to 60 seconds. So 60 minutes will equal to 3600 seconds.

Now we divide 1609.344 by 3600 to find a unit rate of m/s:

[tex]\displaystyle{\dfrac{1609.344}{3600} \ \, \sf{m/s}}\\\\\displaystyle{= 0.44704 \ \, \sf{m/s}}[/tex]

Now multiply 0.44704 m/s by 0 and 60 to get velocity in m/s unit:

  • Initial velocity = 0 m/s
  • Final velocity = 60 * 0.44704 = 26.82 m/s

Time is already in second so no need for conversion. Substitute known information in the formula:

[tex]\displaystyle{\vec{a} = \dfrac{26.82-0}{4-0}}\\\\\displaystyle{\vec{a} = \dfrac{26.82}{4}}\\\\\displaystyle{\vec{a} = 6.7 \ \, \sf{m/s^2}}[/tex]

Therefore, the Firebird will accelerate at the rate of 6.7 m/s^2.

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