Question

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.)

Answer 1

initial speed of the racer is given as

$v_i = 212 mi/h$

$v_i = 212*\frac{1609}{3600} = 94.75 m/s$

after applied force the final speed is given as

$v_f = 45 mi/h$

$v_f = 45 * \frac{1609}{3600} = 20.11 m/s$

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