Note: the net distance along the path is a scalar quantity its direction is not important so the average speed is also a scalar.

Transcription

1 PHY 309 K. Solutions for the first mid-term test /13/014). Problem #1: By definition, aerage speed net distance along the path of motion time. 1) ote: the net distance along the path is a scalar quantity its direction is not important so the aerage speed is also a scalar. The student in question walks along 3 sides of a square mile, so the net distance along his path of motion is 3 miles. The time this motion took was 1 hour, so the aerage speed of the student was 3 miles per hour. On the other hand, the aerage elocity is a ector quantity defined as aerage elocity The net displacement ector of the student is net displacement ector time. ) D net 1 mile South) + 1 mile West) + 1 mile orth) 1 mile West) 3) since the Southward and orthward displacements cancel each other. Consequently, the student s aerage elocity ector is 1 mile per hour in the direction due West. Problem #: a) Since the car has a constant acceleration a, its elocity changes with time according to a ery simple formula t) 0 + at 4) where 0 is the initial elocity. ote that the acceleration a is negatie, so the elocity decreases with time. Gien a and the final elocity, we may find the deceleration time t by simply soling the equation 4) for t: t 0 a 1 m/s) 3 m/s).5 m/s 0 m/s 8.0 s. 5).5 m/s 1

2 b) In class, I showed that when a body has constant acceleration or deceleration), its aerage elocity is the aerage between the initial and the final elocity: ag init + fin ote that this formula applies only when the acceleration is constant! For the car in question, eq. 6) gies its aerage elocity during the period it was decelerating:. 6) ag m/s) + 1 m/s) Consequently, the distance coered by the car during this time is m/s. 7) D t ag 8.0 s m/s 176 m 180 m, 8) or about 00 yards 600 feet). Alternatie solution: Use the equation of motion at constant acceleration, D x x 0 0 t + a t. 9) For the time t we found in part a), this formula gies us D 3m/s) 8.0s)+.5 m/s ) 8.0s) 56m 80m 176m 180m. 10) Problem #3: The elocity ectors of the relatie to the ground and relatie to the moing air differ by the wind elocity ector, gr) + wind, gr) wind. 11) We know the direction and magnitude of the wind elocity ector 9 MPH due West and also direction and magnitude of s elocity ector relatie to the ground 1 MPH

3 due orth. To find the s elocity ector relatie to the air, all we need is to ealuate the ector difference in the second eq. 11). Graphically, wind gr) φ W wind E In,E) components, S gr) 1 MPH, gr) E 0, wind ) 0, wind ) E 9 MPH, 1) hence ) gr) ) wind) 1 MPH 0 MPH +1 MPH, )E gr) )E wind) E 0 MPH 9 MPH) +9 MPH. 13) ow let s conert these components of the s elocity ector relatie to the air into magnitude and direction. The magnitude i.e., the s airspeed obtains from the Pythagoras formula + ) E 1 MPH) +9 MPH) 15 MPH. 14) As to the direction of the i.e., the s heading, we hae tanφ E 9 MPH 1 MPH 0.75, 15) 3

4 hence φ arctan0.75) modulo 180 ) 37 or ) Since the orth and East components of the are both positie, the correct answer is 37. Thus, the in questions has heading 37 clockwise from orth). Problem #4: Once the car exits the suriing part of the bridge, it flies through the air like a projectile until it hits the water. The ertical and the horizontal motion of a projectile are independent: xt) 0x t, yt) y 0 + 0y t 1 gt, 17) where 0 0x, 0y ) is the initial elocity ector of the car at the moment it went off the bridge. The direction of this ector is parallel to the bridge, which is presumably horizontal. Therefore, 0x 0 is the car s speed when it went off the bridge while 0y 0, hence xt) 0 t, 18) yt) y 0 1 gt. 19) a) ote that the ertical motion of the car does not depend on its initial speed but only of the time of its flight. Consequently, we may obtain this time of flight t from the net ertical displacement of the car from the bridge y 0 16 m to the water surface y 0. Thus, soling eq. 19) for the time t, we find y 0 1 gt y 0 g t y 0 t y 0 g t y0 g. 0) For the bridge in question, t 16 m 9.8 m/s 3.7 s 1.8 s. 1) 4

5 b) Eq. 18) relates the horizontal displacement of the falling car to its initial speed 0 and the time of flight t. Gien the known horizontal displacement x 36 m and the flight time t 1.8 s found in part a), the initial speed of the car follows as 0 x t 36 m 1.8 s 0 m/s, ) or about 45 miles per hour. Problem #5: The Second Law of ewton relates the acceleration of the body with the net force acting on it, m a F net F 1 + F +. 3) ote the ector sum here: When adding forces, it s important to know not only their magnitudes but also their directions. The boat in question is subject to two horizontal forces: the propeller force F prop +450 pushing the boat forward, and the water resistance force F res 330 pushing the boat back. The ± signs here indicate the directions of the two forces. In one dimension forward or backward the ector sum of two forces is just the algebraic sum of the two signed numbers. Thus, the net force acting on the boat is F net F prop + F res 450 ) ) +10, 4) where the + sign indicates the forward direction. Gien the net force on the boat, its acceleration follows from the Second Law, a 1 m F net +10 / 40 kg m/s. 5) In other words, the boat accelerates forward at the rate a 0.50 rmm/s. 5