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1 1 PHYS:100 LECTURE 4 MECHANICS (3) Ths lecture covers the eneral case of moton wth constant acceleraton and free fall (whch s one of the more mportant examples of moton wth constant acceleraton) n a more quanttatve manner. The oal here s to dscuss the formulas that can be used to calculate the poston and velocty of an object at any tme f we know the acceleraton and the ntal (bennn) speed of the object. The type of analyss that we wll fnd s the same analyss that s used to reconstruct the facts of an automoble accdent usn some of the nformaton, e.., skd marks that mht be avalable at the scene. We ben by remndn you that acceleraton s the rate of chane of velocty, a v, t where v means the chane n velocty, and t s the tme nterval over whch t chanes. In the scentfc system of unts a s measured n meters per second squared, or m/s. We wll ben by dscussn the smpler stuaton of moton wth constant velocty, or n other words, the case n whch there s no acceleraton, a = Moton wth Constant Velocty. Suppose an object moves wth constant velocty v. Recall that velocty ncludes both speed and drecton, so when we say that the velocty s constant that means that the object s movn wth constant speed n a straht lne nether ts speed or drecton chanes as t moves. To descrbe moton mathematcally, we must frst set up a coordnate system. For moton wth constant velocty, we can take the moton to occur alon the x axs of a coordnate system. The other parameter we must track, of course, s the tme. We can man that we have a vdeo camera and we make a move of the moton of the object. To analyze the moton we nspect the move frame by frame and develop a table of where the object s on each frame of the move. Suppose we call the frst locaton of the partcle x (we read ths as x and t means the ntal poston). There s nothn partcularly specal about x, t s just where we decde to start the analyss, and t can also be taken as the tme when we start the clock,.e., t = 0. Suppose we know that the object s velocty s v. What we want s a formula whch allows us to fure out, where the object wll be at a later (or fnal) tme t call ths poston x f (read x f for x fnal). The formula for x f n terms of x, v, and t s Dstance x f travelled from ntal poston x n tme t when movn wth constant velocty v x = x + vt. [1] f

2 Notce that formula [1] makes sense f v = 0: x f = x, the object stays at ts ntal poston. Now, n settn up the coordnate system, we can smplfy thns and say that the ntal poston s just x = 0. In ths case, the above formula tells us, for example, that f you travel n a straht lne at 75 mph for one hour, you wll travel exactly 75 mles, whch s the answer you would have arrved at usn common sense. 4. Moton wth Constant Acceleraton. Acceleraton s the rate of chane of velocty wth tme. Suppose that at some ntal tme, t 1 an object has a velocty v 1, then at some later tme, t ts velocty s v, then over ths tme nterval ts acceleraton s Δv v v1 Defnton of acceleraton: a = = Δt t t 1. [] Example 4 1: The velocty of an object ncreases from 50 m/s to 75 m/s over an nterval of 5 seconds. What s the acceleraton? Soluton v v 75 ms 50 ms 5 ms t t 5s 5s 1 a = 5. 1 ms Suppose an object moves wth constant acceleraton (a) and at t = 0, ts velocty s v. Its velocty v at a later tme t s computed by Moton wth constant acceleraton v = v + a t [3] In other words, to compute the velocty at a later tme, we add to the ntal velocty the quantty of a tmes t, so that a x t s the amount by whch the velocty ncreases. An example of the applcaton of formula [3] s ven on slde 10. Example 4 : An object movn at 30 m/s bens acceleratn at 5 m/s. When wll ts speed reach 50 m/s? Solutonf v v at 50 ms 30 ms 5 ms t t 5t 0 t 4 s.

3 3 Notce once aan, that formula [3] smply tells us that f a = 0, v = v snce the velocty does not chane. The velocty of an object can also decrease ths s also an example of accelerated moton, but n ths case we use the specal term deceleraton. When you are drvn and apply the brakes, you slow down or decelerate. Deceleraton can be handled n the same way as acceleraton, f we use a mnus sn n front of a. For example a car may slow down f the brakes provde an acceleraton (deceleraton) a = 5 m/s. If you look at formula [3] when a s neatve, the second term s neatve, meann that t subtracts from the frst term so v wll be less than v. Ths s llustrated by the example on slde 11. Example 4 3: A car s movn at 150 m/s when the brakes are appled. If the brakes provde a deceleraton of 30 m/s, how lon wll t take for the car to stop? (unrealstc numbers!) Soluton v v at ( 30) t 30 t 150 t 5 s. f 4 3. Free fall. Free fall s an example of moton wth constant acceleraton. Formula [3] can be appled to free fall by settn a = = 10 m/s. If an object falls from rest (you just let t o) so that v = 0, then ts speed at a later tme t s ven by Velocty after falln for a tme t startn from rest: v = t = 10 t. [4] The table on slde 13 ves the results of observn a falln object by vn ts velocty at one second ntervals and the dstance that t has fallen. You can see that the measured veloctes follow formula [4]. Each second, the velocty ncreases by 10 m/s. The rht hand column of the table on slde 13 ve the dstances fallen each second. The frst pont to notce s that the object falls dfferent dstances n each successve second of free fall: 5 m for the frst second, 15 addtonal m durn the second second, 5 m durn the thrd second, etc. A freely falln object does not fall the same dstance n each successve tme nterval. The numbers n the rht column are also expressed n a form that llustrates a pattern the vertcal dstance travelled

4 4 durn each second appears to be ven by ½ (10) x the square of the tme. Galleo was the frst to notce ths pattern and was able to wrte down the formula for the dstance (d) an object falls at any tme, t assumn t was dropped from rest: The dstance y that an object falls n a tme t, startn from rest: y = 1 t [5] An example of the use of formula [5] s ven on slde 17. Formulas [4] and [5] can be used to fure out how lon t would take for an object to fall a partcular dstance, and how fast t would be movn when t ht the round. An example s ven on sldes 18 and 19. Example 4 4: An object falls from rest from a reat heht. How fast wll t be movn and how far wll t have fallen n 10 seconds? Soluton 1 1 v t m s s m s d t m s s m 10 / (10 ) 100 / ; 10 / (10 ) 5(100) 500. Formula [5] can also be used to analyze the problem of an object that s thrown straht up and then comes down. To throw an object up means that t s ven an ntal upward velocty, v. As t rses, ravty causes t velocty to decrease to zero at the top of ts path, then ravty causes t to fall back down, ncreasn ts velocty on the way down. In fact, n the absence of ar resstance, t turns out that when the object falls back to ts ornal poston, t wll have the same speed that t was ven when t was tossed up. Note that t has the same speed, not the same velocty, because the drecton of ts velocty s not the same. Ths problem can be understood on the bass of equaton [3]. Consder the frst part of the moton where the object s thrown upward. If we consder the upward drecton to be postve, then v s postve and the acceleraton s neatve (a = = 10 m/s ). At the very top of the object s path, v = 0, so equaton [3] says that: 0 = v t, or v = t, or the tme that t takes for an object thrown upward wth a speed v to reach ts hhest pont s Tme to reach hhest pont startn wth v : v t up =. [6]

5 5 Now, t takes exactly the same amount of tme to rse to ts hhest pont as t takes to fall back to the round, So the total tme the object s n the ar s t total = t up + t down, and snce t up = t down the total tme the object s n the ar s Tme for an object thrown up wth speed v to return back down: v t total =. [8] Example 4.5: Lebron James jumps straht up wth a speed of 5 m/s to make a jump shot. How lon s he n the ar? Soluton What we are lookn for s the total tme he s n the ar, whch s the tme to jump up v0 5 m/ s plus the tme to fall back down. The tme to jump up s t up 0.5 s. 10 m/ s Snce the tme to jump up and the tme to fall back down are the same, the total tme he wll be n the ar s 1 s.

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