Question

INITIAL DATA THE WEIGHT OF THE CHUPACABRA THAT YOU WILL USE IN THIS PROBLEM IS DETERMINED IN THIS WAY CHUPACABRA WEIGHT = (YOUR WEIGHT IN POUNDS x 0.5) + (YOUR HEIGHT IN INCHES * 0.5) + (LAST 2 DIGITS OF YOUR STUDENT NUMBER) EXAMPLE IN MY CASE I WEIGHT 300 POUNDS, MEASURE 5 FEET 10 INCHES (70 INCHES) AND MY #EST IS 802-XX-0804 CHUPACABRA WEIGHT W = (300 POUNDS OF WEIGHT X 0.5) + (70 INCHES * 0.5) + (04) = 189 lbf DESCRIPTION OF THE PROBLEM The chupacabra (of weight W) runs his skateboard at a speed V = 49 feet per second (figure 1). The chupacabra detects barriers and applies the brakes for a distance D before = 21 feet. Then it collides with barrier 1 long D1 = 11.2 feet and exerts a force F1 = 88 pounds and then with barrier 2 long D2 = 38.2 feet that exerts a force F2 = 59 pounds. The surface has a coefficient of friction m = 0.29 THE FIRST QUESTION IS ANSWERED IN MOODLE, THE REST IN PDF 1. Determine the speed of the chupacabra before touching the barrier 1 2. Determine the speed of the chupacabra before touching barrier 2 3. Can barriers stop the chupacabra? and if they can, where does the chupacabra end?

Answer 1

# Student has not provided with his weight. So will assume he weighs 150 lb # height = 5 ft 10 inches = 70 inches # last two digits of Roll No: = 50 [ You have to repeat below Calculations again by putting your own actual values of weight, height and last Awo digits of your Roll No.] weight of =(150x0.5) + (70x0.5) 750 Chupakabra = 160 lbf

# Force during of all friction which acts time of travel = M N = urma) = (0.29) x 160 = 46.4 I bf • # Before touching Barrier 1 • Distance covered CD 1= 21 ff Resistive acceleration = Force of friction Mass of Chupakabra = 46.4 lbf) ==9.3305 fut (160 ÷ 32.174) Ibm sec²

• Initial Velocity of Chupakabra = 49 ft/sec * Using third law of kinematics. 22=u² +2 as - Substituting Values, : K2= (4992 + 26–9.3305)(21) i v = 44.823 2 ft/sec .: Speed of Chupakabra before touching Barrier 1 = 44.8232 ff # Before touching Barrier 2 sec - - A constant Resistive force of FI=88 e bf acts over a distance

D,-C11.2 ft long). Total Resistive [ Fit friction I 88 + 46.4 = 134.4 lbf Force force 1 ☆ Resis Hive acceleration i Total Resistive Force mass of Chupa kabra = 134.4 lbf (160 = 32.174) ebm = -2 7.0262 fit seca Initial velocity before Barries I = 44.82 32 ft/sec

* Using third law of kinematics: 22 = u²+zas Substituting Values, u? = (44.82327 +26-27.0262) (11.2) x = 37.4664 ft - sec ... Speed of Chupakabra before touching Barrier 2= 37.4664 ft sec # Now, chupakabra has crossed Barrier I and now it will Collide to Barrier 2. -A constant Resistive force of z = 59 ebf acts over a distance

D2 = 38.2 fut Force force Total Resistive [ Ęt friction ī 59 + 46.4 = 105.4 lbf - - ☆ Resistive acceleration i Total Resistive Force Mass of Chupa kabra 105.4 lbf 2.21.1946 fut (160 - 32.174) ebm z sec? velocity before colliding ... Initial Barrier 2 = 37.4664 fit sec

* Using third law of kinematics: - vz=u² +z as substituting Values, : *2= (37.466432 +26-211946)(382) - v= -215.54 → This negative value implies that Chupa Kabra has been completely stopped in between Barrier 2. Hence, Barrier 2 did stop chupa kabra. Let us find out at what distance

Now, we know, final velocity (12)=0 * Using third law of kinematics. 2²=u²+zas Substituting Values, :(032 = (37.466477 26–21.194095 Solving, s= 33.1153 fut ☆ Hence, Chupakabra has been stopped 33.1153 fut after collision with barrier 2 * From initial location, he was stopped after (D+ D, +33-11) = 65.3153 ft