Question

1. A car accelerates from zero to 105 km/h in 5.21 s. What is its average acceleration in km/h/s and m/s? 2. An object moving in the x-direction has a position described by the function x(t) - 6.30t + 8.70t2, where t is in seconds and x is in meters. What is the acceleration as a function of time? The figure shows the graph of v vs t for the motion of a motorcyle that starts from rest and moves along the road in a straight line 3. a. Find the average acceleration for the time interval t 0 to t 6.00 s Estimate the time at which the acceleration has its greatest positive value and the value of the acceleration at that instant When is the acceleration zero? Estimate the maximum negative acceleration and the time at which it occurs b. c. d. 10 12 14 Starting from rest, a light plane must reach a speed of 130 mi/h in order to take off. How long a runway is needed if the plane can accelerate at a constant 3.50 m/s? 4. A car traveling at 35 mi/h crashes into a tree. The front end of the car compresses and the driver comes to rest after traveling 0.75 m. What was the magnitude of the average acceleration of the driver during the collision? Express your answer in terms of "g's", where 1 g 9.80 m/s2. Determine the time interval of the collision, in milliseconds 5. The tower of Pisa is 54.5 m tall. Assuming that Galileo dropped his object from rest from the top of the tower and that the effects of air resistance were negligible, how long would it take the object to fall, in seconds? What is the velocity of the object when it hits the ground? 6. 7. A ball is thrown directly downward with an initial speed of 9.50 m/s from a height of 30.0 m. After what time interval does the ball strike the ground? What is its velocity as it reaches the ground?

Answer 1

Acceleration (a) = v_2 - v_1/t_2 - t_1 = 105 - 0/5.21 - 0 km/h/sec = 20.15 km/h/sec OR, v_2 = 105 km/hr = 105 times 1000/3600 m/sec = 175/6 m/s^2 = 29.17 m/s Therefore acceleration (a) = 29.17 - 0/5.21 - 0 m/s^2 = 5.59 m/sec^2 \r\n Position as a function of time x(t) = 6.30t + 8.70t^2 or, velocity v = dx/dt = 6.30 + 8.70 times 2t = 6.30 + 17.4t Therefore acceleration (a) = d^2 x/dt^2 = 17.4 m/sec^2 Final speed (v_f) = 130 m/hr = 130 times 1609/3600 m/sec = 58.10 m/sec From v^2 = u^2 + 2as or, (58.10)^2 = 0 + 2 times 3.50 times s or, s = 482.28 m = length of runway needed From v^2 = u^2 + 2gh or, u^2 = (9.50)^2 + 2 times 9.8 times 30 or, u = 26.04 m/sec = velocity of ball as it reaches ground From v = u + at