If you are driving 72 Km/h along a straight road and you reduce your speed to 45 km/h in 4.0 s,what is your acceleration during this inattentive period?

Respuesta :

Answer:

[tex]a=-1.875\ m/s^2[/tex]

Explanation:

Motion With Constant Acceleration

It occurs when the velocity of an object changes uniformly in time.

The equation that rules the change of velocities is:

[tex]v_f=v_o+at[/tex]

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

Solving for a:

[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]

The initial speed is vo=72 Km/h and it's reduced to vf=45 km/h in a time of t=4 s. Both speeds will be converted to m/s:

vo=72*1000/3600 = 20 m/s

vf=45*1000/3600 = 12.5 m/s

Now calculate the acceleration:

[tex]\displaystyle a=\frac{12.5-20}{4}=\frac{-7.5}{4}[/tex]

[tex]\mathbf{a=-1.875\ m/s^2}[/tex]

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