Take g-10m/s^-2Give answers to 2 significant figures.Formulas and calculations required for all answers.A racing bike is being put through its paces on a straight stretch of road by a stunt man. The following velocity-time graph was obtained upon analysis of the bikes performance characteristics: 1. a) Calculate the distance travelled by the bike during the first 40 seconds.To calculate the distance, it equals the area under the graph from 0sec to 40 sec. The area is made of 2 trapezoids, one from 0sec to 30sec and the other from 30sec to 40sec.

Area of a trapezoid: (base 1 + base 2) x height 0-30sec Trapezoid = (10 + 30) x 60 = 1200m 30-40sec Trapezoid = (40 + 60) x 10 = 500m

Total area = area of 2 trapezoids = 1200 + 500 = 1700m

b) What was the acceleration of the bike after it started from rest?(i) 5seconds Acceleration = v/t = slope of the velocity-time graph = slope of the graph for the first 20sec, which is positive = (60m/s 0 m/s)/(20sec 0sec) = 3.0 m/s2 positive acceleration(ii) 22secondsAcceleration = v/t = slope of the velocity-time graph = slope of the graph from 20-30sec, which is zero = (60m/s 60m/s)/(30sec 20sec) = 0 m/s2 constant velocity, so no acceleration(iii) 50secondsAcceleration = v/t = slope of the velocity-time graph = slope of the graph from 30-60sec, which is negative = -2.0m/s2 negative acceleration

2. On the Pentecost Islands there is an initiation rite called land diving. The islanders jump from a high platform with a rope made of vines attached to their ankles. The ropes are designed to bring the land divers to a stop just centimeters above the ground after they have fallen.a) Calculate the velocity of a land diver assuming he starts from rest, falls 25meters, and that friction is negligible.v = 2gdv = 2*(10m/s2)*(25m)v = 22 m/sb) How long does it take for the diver to fall the 25meters under the influence of gravity?t = 2d/gt = 2*(25m)/(10m/s2)t = 2.2secAfter travelling 25meters the vines attached to the land divers ankles start to stretch and cause the land diver to decelerate and come to a stop. The vine stretches by 50centimeters.c) Calculate the magnitude of the deceleration of the land diver whilst stopping.vf2 = vi2 + 2advf = 0m/svi = 22m/sd = 0.5m

0 = (22m/s)2 + 2a(0.5m)-484m2/s2 = 1m*(a)-484m/s2 = aor 484m/s2 downward3. A block on the end of a string 3meters long is swung in an anticlockwise horizontal circle with a constant speed of 5m/s. You are looking down (birds-eye view) on to this horizontal circle.

a) On the diagram above draw, using arrows, the instantaneous velocity of the block at the South (S) position and at the East (E) position.See diagram.b) What is the change in velocity (magnitude and direction) of the block as its motion changes from the south (S) position to the East (E) position in its circular path?v =(5m/s)2 + (5m/s)2 = 7.1m/s towards the center of the circleSee diagram.

4. A car of mass 800kg accelerates from rest to 20m/s in 8.0s. The resistance forces acting on the car total 1000N.a) Draw a well labeled diagram showing all forces acting on the car. Clearly show the point of application of each force, ie, show where each force starts.See diagramb) Calculate the driving force being provided by the engine of the car.Acceleration (a) = v/t = (20m/s 0m/s)/(8sec) = 2.5m/s2

F = FD FR = ma, where FD is the driving force and FR is the resistive force. FD 1000N = (800kg)(2.5m/s2) FD = 2000N + 1000N FD = 3000N

5. A water-skier of mass 70kg is accelerating at 2.5 m/s2. The tension T in the rope due to the boat that pulls the skier is 200N. See over page.

a) On the diagram above, draw and label two arrows that show the action-reaction forces acting on the rope between the boat and the skier. Use the letters A for action, and R for reaction next to your arrows.See diagram.b) Complete the two sentences below, which describe the action-reaction pair above.Action: The force of the boat pulling on the skierReaction: The force of the skier pulling on the boatc) Calculate the magnitude (size) of the drag (resistance) forces acting on the skier.FD = T ma = 200N (70kg)(2.5m/s2) = 25 N

6.A toy car of mass 50grams travels down a smooth incline at 30 degrees to the horizontal. Friction may be ignored.a) Calculate the net force acting on the car as it rolls down the slope.Sum of the forces= F= ma = mg (sin) = (0.05kg)*(10m/s2)*(sin(30)) = 0.25NSee diagram.b) Calculate the force of the incline on the car as it travels down the slope.The force of the incline on the car is perpendicular to the slope of the incline and it is the normal force, FN.See diagram.FN = mg (cos) = (0.05kg)*(10m/s2)*(cos(30)) = 0.43N

7.A car drives off a 40m high cliff at 20m/s horizontally as shown below.

a) How long does the car take to land?See diagram.x = ?vxi = 20m/svxf = 20m/sax = 0 m/s2y = 40mvyi = 0m/svyf = ?ay = -10m/s2

Now calculate the time.y = vyi(t) + ay(t2)-40m = 0(t) + (-10m/s2)(t2)-40m = -5m/s2(t2)8s2 = t22.8s = tb) How far does it land from the base of the cliff?Using the table from above and time calculated. Solve for x.x = vxi(t) + ax(t2) = (20m/s)(2.8s) + (0m/s2)(2.8s)2 = 56mc) Using vertical and horizontal velocity components, what is the cars velocity just before it lands? State the magnitude and direction(Use formula(s), calculations and a vector diagram)See diagram.Horizontal velocity component = 20m/s, rightVertical velocity component = 10m/s, down for each secondVertical velocity after 2.8s => vf = vi + a(t) vf = 0 + (-10m/s2)(2.8s)= -28m/s or 28m/s downward

The cars velocity after 2.8s:v* = vx2 + vy2 = (20m/s)2 + (28m/s)2 = 34m/s Direction of the cars motion after 2.8s: tan = vy/vx = 34/20 ---- > = 60 to the horizontalFinal answer: 34m/s downward or 60 to the horizontal

8.A catapult launches a marble, of mass 100grams at 25m/s at an angle of 30 degrees above the horizontal.

a) What is the maximum height reached by the marble?Draw a right triangle with the smallest angle = 30. Label the hypotenuse 25m/s and the side opposite the angle vyi.To solve for vyi:sin 30 = opposite/hypotenuse = vyi/(25m/s)vyi = 12.5m/s

To calculate the height, we used the same right triangle:(vyf)2 = (vyi)2 + 2adydy = (vyf2 vyi2)/2ady = (02 12.52)/2(-10m/s2)dy = 7.8mb) What is the time of flight of the marble?x = ?vxi = 22m/svxf = 22m/sax = 0 m/s2y = 7.8m (max height)vyi = 12.5m/svyf = ?ay = -10m/s2

y = vyi(t) + a(t2)0 = (12.5m/s)(t) + (-10m/s2)(t2)-12.5m/s = (-5m/s2)(t)2.5s = tWhat is the magnitude and direction of the accelerations of the marble at:(Answers + one dot-point explanation required)(i)X => horizontal acceleration (ax) = 0m/s2 Right; vertical acceleration (ay) = 10m/s2 Down(ii)Y => horizontal acceleration (ax) = 0m/s2 Right; vertical acceleration (ay) = 10m/s2 Down(iii)Z => horizontal acceleration (ax) = 0m/s2 Right; vertical acceleration (ay) = 10m/s2 Down

There is no horizontal acceleration for all three points because the horizontal velocity of the marble is constant and because there is no horizontal force, the acceleration is zero. The vertical acceleration for any projectile including this one is always 10m/s2, downward due to the force of gravity.

9.A tennis ball of mass 100g hits the wall horizontally at 8.0m/s east and rebounds at 6.0m/s. The contact time with the wall is 0.07s.a) Calculate the impulse on the ball by the wall.mball = 0.1kgvball(i) = 8m/s Eastvball(f) = 6m/s WestMomentum (p) = m x v, where m = mass and v = velocity Impulse (I) = p = mball(vball(f)) mball(vball(i)) = (0.1kg) (-6m/s) (0.1kg)(8m/s) = -1.4kg(m/s) = 1.4 Ns Westb) Calculate the change in momentum of the ball.p = mball(vball(f)) mball(vball(i)) = (0.1kg) (-6m/s) (0.1kg)(8m/s) = -1.4kg(m/s) = 1.4kg(m/s) West

See diagram.State the magnitude and direction for a) and b)A vector diagram, formula(s) and calculations required.10.Calculate:a) The force exerted on the ball by the wallF = m x a = p/tF = -1.4 kg(m/s)/0.07sF = -20NFwall on ball = 20N Westb) The force exerted on the wall by the ballFball on wall = 20N EastBecause there is an equal and opposite reaction to every action, the ball exacts the same amount of force in the opposite direction. State the magnitude and direction for a) and b)Formulas and calculations required.