Hw 2: Motion in 1D and Vectors Velocity 2.1) [2.13 in text book] A person takes a trip, driving with a constant speed of 89.5 km/h, except for a 22.0 minute rest stop. If the persons average speed is 77.8 km/h, a) How much time is spent on the trip? [2.80 hr] b) How far does the person travel? [218 km] Motion under constant acceleration 2.2) A car on the highway slows from 32.0 m/s to rest as the driver approaches a traffic jam. The car has 70.0 m to stop. a) Solve for the acceleration needed to stop the car. [-7.31 m/s2] b) Solve for the time needed to stop. [4.37 s] Free Fallin' 2.3) [2.46 in text book] A ball is thrown directly downward with an initial speed of 8.00 m/s, from a height of 30.0 m. After what time interval does it strike the ground? [1.79 s] 2.4) [2.53 in text book] A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s2 until its engines stop at an altitude of 150 m. a) What can you say about the motion of the rocket after its engines stop? [t=2.83s and v=55.67m/s] b) What is the maximum height reached by the rocket? [308 m] c) How long after liftoff does the rocket reach its maximum height? [8.51 s] d) How long is the rocket in the air? [16.4 s] Vectors 2.5) [3.14 in text book] A hiker starts at his camp and move the following distances while exploring his surroundings: 75.0 m north, 2.50102 m east, 125 m at an angle of 30.0north of east, and 1.50 102 m south. a) Find his resultant displacement from camp. Take east as the positive x-direction and north as the positive y-direction. [~R = 358.25 m,-12.5 m)] b) Would changes in the order in which the hiker makes the given displacements alter his final position? 2.6) [3.20 in text book] The helicopter view of the figure below shows two people pulling on a stubborn mule. Find: a) the single force that is equivalent to the two forces shown. (|~F1|= 80.0 N and |~F2|= 120 N ) [~R = (39.3 N,181 N)] b) the force a third person would have to exert on the mule to make the net force equal to zero. The forces are measured in units of Newtons (N ). 1

Help on Problem 1:

Break the problem up into two parts: Part A where the car is driving and part B where the car is stopped

Part A

xA=?tA=?vA=89.5kmhr

Part B

xB=0mtB=0.366hrvB=0kmhr

Total Trip

xT=xA+xBtT=tA+tBvT=77.8kmhr

System of equations:

Now we need to identify a physical equation that is of use in this system. First of all, convince yourself that there is not acceleration for either part of the trip (constant velocity). Using the definition of velocity (v=x/t), you are able to write the following equations:

vA=xAtAvT=xTtT=xA+xBtA+tB

Substituting values into these equations yields:

89.5=xAtA77.8=xAtA+0.366

This is 2 equations and 2 unknowns. Solve the system of equations to find the final answers to the question.