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A car is traveling in a straight line path at a maximum speed of 7.00 m/s. The driver of the vehicle applies the brakes bringing the car to a stop after traveling 10.0 m. What is the magnitude of the car's acceleration to the nearest hundredths place?

Answer :

elcharly64

Answer:

[tex]a=-2.45\ m/s^2[/tex]

Explanation:

An object is traveling at constant acceleration if its changes in speed are constant at the same intervals of time.

The acceleration can be calculated as follows:

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

Where vo is the initial speed, vf the final speed and t is the time.

The acceleration and the distance x are related through the following equation:

[tex]v_f^2=v_o^2+2.a.x[/tex]

The car has an initial speed of vo=7 m/s when the driver stops the car (vf=0) after traveling x=10 m.

The acceleration can be calculated by solving the last equation for a:

[tex]\displaystyle a=\frac{v_f^2-v_o^2}{2x}[/tex]

[tex]\displaystyle a=\frac{0-7^2}{2*10}[/tex]

[tex]\displaystyle a=\frac{-49}{20}[/tex]

[tex]\boxed{a=-2.45\ m/s^2}[/tex]

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