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A drag racer crosses the finish line doing 212 mi/h and promptly deploys her drag chute. (A.) what force must the drag chute exert on the 898- kg car to slow it to 45 mi/h in a distance of 185m? (B.) describe the strategy you used to solve (a.)

1 Answer

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initial speed of the racer is given as


v_i = 212 mi/h


v_i = 212*(1609)/(3600) = 94.75 m/s

after applied force the final speed is given as


v_f = 45 mi/h


v_f = 45 * (1609)/(3600) = 20.11 m/s

now during this speed change the racer will cover total distance 185 m

so here we will use kinematics


v_f^2 - v_i^2 = 2 a d


20.11^2 - 94.75^2 = 2*a*185


a = -23.2 m/s^2

now the force that chute will exert on the racer will be given as


F = ma


F = 898* 23.2


F = 2.1* 10^4 N

B) here following is the strategy for solving it

1. first we used kinematics to find the acceleration of the car

2. then we used Newton's II law (F = ma) to find the force

User Krishna Srinivas
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