Tutorial 01 2018 ans PDF

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64-130 Tutorial 01

Problems : 11. Ch 2, Problem 23. 2 3 . T womo t o r c y c l e sa r et r a v e l i n gd uee a s twi t hd i ffe r e n tv e l o c i t i e s . Ho we v e r ,f o u rs e c o n d sl a t e r ,t he yh a v e t hes a mev e l o c i t y . Du r i n gt hi sf o u r s e c o n di n t e r v a l , c y c l eAh a sa na v e r a g ea c c e l e r a t i o nof d u ee a s t ,whi l ec y c l eBh a sa na v e r a g ea c c e l e r a t i o nof d u ee a s t . Byh o wmuc hd i dt h e s p e e dsd i ffe ra tt heb e g i n n i n go ft h ef o u r s e c o n di n t e r v a l ,a n dwh i c hmo t o r c y c l ewa smo v i n gf a s t e r ?

REASONING AND SOLUTION Both motorcycles have the same velocity v at the end of the four second interval. Now v=v0A +a t A for motorcycle A and t v=v +a 0 B B for motorcycle B. Subtraction of these equations and rearrangement gives 2

2

v –v0B=( 4. 0m/ s –2. 0m/ s) ( 4s )= 0 A

8.0 m/s

The positive result indicates that motorcycle A was initially traveling faster.

12. . Ch 2, Problem 30 3 0. TheKe nt u c kyDe r b yi she l da tt heChu r c hi l lDo wnst r a c ki nLou i s v i l l e , Ke nt uc k y . Thet r a c k i so nea ndo ne q ua r t e rmi l e si nl e n g t h. Oneo ft hemos tf a moushor s e st owi nt hi se v e n twa s Se c r e t a r i a t .I n1 973hes e taDe r b yr e c or dt ha two ul dbeh a r dt obe a t .Hi sa v e r a g e a c c e l e r a t i ond ur i n gt hel a s tf ourqu a r t e r mi l e soft her a c ewa s

. Hi sv e l oc i t y

a tt hes t a r to ft hefina lmi l e wa sa b out . Thea c c e l e r a t i on, a l t hou ghs ma l l , wa sv e r yi mp or t a ntt ohi sv i c t or y . Toa s s e s si t se ffe c t , de t e r mi net he d i ffe r e n c ebe t we e nt het i mehewou l dha v et a k e nt or unt hefina lmi l ea tac on s t a ntv e l oc i t y o f a ndt het i mehea c t u a l l yt ook . Al t hou g ht h et r a c ki so v a li ns ha pe , a s s ume i ti ss t r a i g htf ort hepu r po s eoft hi spr ob l e m.

REASONING At a constant velocity the time required for Secretariat to run the final mile is given by Equation 2.2 as the displacement (+1609 m) divided by the velocity. The actual time required for Secretariat to run the final mile can be determined from

Equation 2.8, since the initial velocity, the acceleration, and the displacement are given. It is the difference between these two results for the time that we seek. SOLUTION According to Equation 2.2, with the assumption that the initial time is t0 = 0 s, the run time at a constant velocity is t t  t 0 t 

x 1609 m  97.04 s v 16.58 m/s

x v0t  1 at 2   2 Solving Equation 2.8

t

 21 a    x  2  12 a 

 v0  v02  4

 16.58 m/s   16.58 m/s   4 2



for the time shows that

2

 12   0.0105 m/s 2    1609 m 

 12   0.0105 m/s2 

94.2 s

We have ignored the negative root as being unphysical. The acceleration allowed Secretariat to run the last mile in a time that was faster by 97.04 s  94.2 s  2.8 s

________________________________________________________________________ ___ 13.Ch.2, Problem 34 Ar a c ed r i v e rha sma deapi ts t opt or e f ue l .Af t e rr e f ue l i n g ,hes t a r t sf r omr e s ta ndl e a v e st hepi t a r e awi t ha na c c e l e r a t i o nwh os ema gni t ud ei s6. 0m/s2;a f t e r4. 0 s he e n t e r st h ema i ns pe e d wa y . Att h es a mei ns t a nt ,a not he rc a ront h es pe e d wa ya ndt r a v e l i nga tac ons t a ntv e l oc i t yo f70. 0m/ s o v e r t a ke sa ndpa s s e st hee n t e r i n gc a r .Th ee nt e r i n gc a rma i n t a i nsi t sa c c e l e r a t i on .Ho wmuc ht i me i sr e q ui r e df ort hee n t e r i n gc a rt oc a t c ht heo t he rc a r ?

REASONING The entering car maintains a constant acceleration of a1 = 6.0 m/s 2 from the time it starts from rest in the pit area until it catches the other car, but it is convenient to separate its motion into two intervals. During the first interval, lasting t1 = 4.0 s, it accelerates from rest to the velocity v01 with which it enters the main speedway. This velocity is found from Equation 2.4 = a1, t = t1, and v = v01:

 v v0  at , with v

0

v10 a1t1

= 0 m/s, a

(1)

The second interval begins when the entering car enters the main speedway with velocity v01, and ends when it catches up with the other car, which travels with a constant velocity v02 = 70.0 m/s. Since both cars begin and end the interval side-byside, they both undergo the same displacement x during this interval. The x v0t  12 at 2 displacement of each car is given by Equation 2.8 . For the accelerating car, v0 = v10, and a = a1, so



x v10t  12 a1t 2



(2)

For the other car, v0 = v02 and a = 0 m/s2, and so Equation 2.8 yields x v20 t

(3)

SOLUTION The displacement during the second interval is not required, so equating the right hand sides of Equations (2) and (3) eliminates x, leaving an equation that may be solved for the elapsed time t, which is now the only unknown quantity: v10t  12 a1t2 v20t v10 t  12 a1t 1at 2 1

t

2

v20 t

v 20  v 10

2 v20  v10  a1

Substituting Equation (1) for v10 into Equation (4), we find that

(4)

t

2 v20  a1t1  a1





2  70.0 m/s  6.0 m/s 2  4.0 s     15 s   2 6.0 m/s





14. Ch2. Problem 41. A locomotive is accelerating at 1.6 m/s2. It passes through a 20.0-m-wide crossing in a time of 2.4 s. After the locomotive leaves the crossing, how much time is required until its speed reaches 32 m/s?

REASONING As the train passes through the crossing, its motion is described by  x  1 ( v  v )t  0  2 Equations 2.4 (v = v0 + at) and 2.7  , which can be rearranged to give v  v0 at

a nd

v  v0 

2x t

These can be solved simultaneously to obtain the speed v when the train reaches the end of the crossing. Once v is known, Equation 2.4 can be used to find the time required for the train to reach a speed of 32 m/s. SOLUTION Adding the above equations and solving for v, we obtain 2x  2(20.0 m)    v  12  at    12  (1.6 m/s 2 )(2.4 s)  1.0 101 m/s  t  2.4 s   

The motion from the end of the crossing until the locomotive reaches a speed of 32 m/s requires a time v  v0 32 m/s  1.0 10 1 m/s   14 s t a 1.6 m/s 2...