Athief is trying to escape from a parking garage after completing a robbery, and the thief's car is speeding (v = 13 m/s) toward the door of the parking garage (fig. p2.60). when the thief is l = 22 m from the door a police officer flips a switch to close the garage door. the door starts at a height of 9 m and moves downward at 0.6 m/s. if the thief's car is 1.4 m tall, will the thief escape?

Answer

Yes, the car reaches the door before the gate closes.

Explanation

The time taken by the car to reach at the door.

Time = distance / time

        = 22/13

          = 1.6923 seconds

Time taken by the door to close up to the height of the car.

Distance the door has to move to prevent the car from escaping = 9.1.4 = 7.6 m

From newton's 2nd law of motion;

s = ut + 1/2 gt²

7.6 = 0.6t + 1/2 × 10t²

7.6 = 0.6t + 5t²

50t² + 6t - 76 = 0

Solving this quadrilatic equation,

t = 28.537 seconds

Yes, the car reaches the door before the gate closes.

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft