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3 years ago

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A car travels along a highway with a velocity of 24 m/s, west. The car exits the highway; and 4.0 s later, its instantaneous vel

ocity is 16 m/s, 45° north of west. What is the magnitude of the average acceleration of the car during the four-second interval?

1 answer:

zloy xaker [14]3 years ago

5 0

Answer:

$4.25 m/s^{2}$

Explanation:

Change in velocity considering the x component will be

Final velocity-Initial velocity

$\triangle v_x= 16cos 45^{\circ}-24=-12.6862915 m/s$

Change in velocity considering the y component will be

Final velocity-Initial velocity

$\triangle v_y= 16sin 45^{\circ}-0=11.3137085 m/s$

Resultant change in velocity $=\sqrt {(-12.6862915 m/s)^{2}+(11.3137085 m/s)^{2}}=16.9982938 m/s$

Acceleration= change in velocity per unit time hence

$a= \frac {16.9982938}{4}=4.24957345\approx 4.25 m/s^{2}$

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Answer:

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Explanation:

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Answer:

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Explanation:

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Magnitude of charges , $q=1\ nC = 10^{-9}\ C.$

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