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Question

# You are traveling north and make a $90^\circ$ right -hand turn east on a flat road while driving a car that has a total weight of $3600\ \mathrm{lb}$. Before the turn, the car was traveling at $40\ \mathrm{mi/ h}$, and after the turn is completed you have slowed to $30\ \mathrm{mi/ h}$. If the turn took $4.25\ \mathrm{s}$ to complete, determine the following : (a) the car 's change in kinetic energy, (b) the car 's change in momentum (including direction), and (c) the average net force exerted on the car during the turn (including direction) .

Solution

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Step 1
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Given:

• Mass of the car: $m = 3600 \mathrm{~lb}$;
• Velocity of the car to the north: $v_1 = 40\mathrm{\frac{~mi}{~h}} \hat{y}$;
• Velocity of the car to the east: $v_2 = 30\mathrm{\frac{~mi}{~h}} \hat{x}$;
• Time needed for the turn: $t = 4.25 \mathrm{~s}$

Required: a) The change in car's kinetic energy $\Delta K$; b) The change in car's momentum $\Delta \vec p$; c) The average net force on the car during the turn $\vec F_{avg}$:

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