Practice Exam (Chap. 1, 2, 3) 1. An object is moving only along the x-axis. Its velocity is given by 23()vt t t tαβ γ=+ + where α, β and γ are known constants. (a) What is the object’s acceleration? (solution) 2()() 2 3dv tat t tdtαβ γ==++ (b) How far does it travel between the times 1t = sec and 2 sec?t = (solution) ()2223 2 3 41122 33 441112341113715(2 1 ) (2 1 ) (2 1 )234234ttxtttdttttαβ γ α β γαβγ αβλ==∆ =++=++= − + − + − =++∫ 2. (a) The equation of motion for an object is given by 32() ( )xtbtcte f=+− + where x is a position, t is the time, and b,c,e and f are constants. Obtain its velocity ()vt. (Hint: 1nndtntdt−=.) (solution) 2()() 3 2( )dx tvt bt ct edt==+−(b) The acceleration of an object which is moving along a straight line is given by 5() 3 4at t t=+ What is its position()xt? Assume that the object is initially(at 0)t =at rest at the origin, i.e. (0) 0x =. (solution) 5262626 3 73734() (3 4 )2632() since (0) 02332 1 2() ()23 22112() since (0) 0221vt t t dt C t t Cvt t t vxtvtdt ttdtttCxt t t x=+ +=++∴ =+ =⎛⎞⎟⎜==+=++⎟⎜⎟⎜⎝⎠∴ =+ =∫∫∫ 3. A person initially at point P in the illustration stays there for a moment and then moves along the axis to Q with velocity 1v and stays there for a moment. Then she runs quickly to R with velocity2v , stays there for a moment, and then strolls slowly back to P with velocity3v . (Therefore, 312vvv<<) 01234QRP Draw the position vs. time graph of this motion.4. A football is punted with an initial v0 at an angle of α above the horizontal on a long flat football field. Assume that the football leaves the punter’s foot at the ground level and calculate the following. (a) The maximum height reached by the football. (b) The total time in the air before the ball hits the ground. (c) The total horizontal distance covered, i.e. the range. (d) The velocity of the football at the maximum height. (solution) 0020000220020cos , ( cos )1sin , ( sin )2( ) condition: 0sinsin 0 ,sin()2()maximum time 2sin() 0( ) range 2sincos()( ) at the maximum height,xyyaaavv xv tvv gtyv t gtavvvgt tgvHytgbTvyT TgcRvRxTgdαααααααααα=== − = −=− = ⇒ ==== ⇒ ===000, cosyxxvvvvα===5. (solution)6. In a uniform circular motion of a ball A, the centripetal acceleration is 2224AAAAAvRaRTπ== , where AR is the radius of the circular path, Av is the velocity of the ball A, and AT is its period. (a) What is the definition of the period T? (solution) The period is the time for one revolution (one complete trip around the circle). (b) Another ball B is doing a uniform circular motion having the radius of RB =2RA. Assume that the centripetal acceleration of this ball is the same as the ball A. What is the velocity of the ball B? What is the period of the ball B? Write the answers in terms of the corresponding quantities of the ball A. (solution) 222, 24/2, 24/BBBBBAAAAABBBBBAAAAARavRvvvRRaRaTTTTTTRaππ===
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