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Ch-2

Ch-2. Help-Session. T072 Q2.The position of an object is given as a function of time by , x= 4t 2 -3t 3 ; where x is in meters and t is in seconds. Its average acceleration during the interval from t = 1.0 s to t = 2.0 s is: (Ans: −19 m/s2).

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Ch-2

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  1. Ch-2 Help-Session

  2. T072 Q2.The position of an object is given as a function of time by , x= 4t2-3t3 ; where x is in meters and t is in seconds. Its average acceleration during the interval from t = 1.0 s to t = 2.0 s is: (Ans: −19 m/s2) Q3.: A car starts from rest and undergoes a constant acceleration. It travels 5.0 m in the time interval from t = 0 to t = 1.0 s. Find the displacement of the car during the time interval from t = 1.0 s to t = 2.0 s.( Ans: 15 m) CH-2-072 x1=vit+at2/2 vavg= (vi+vf)/2= vf/2=x/t = vf/2=x/t=5; vf=2x5=10 m/s a =vf-vi/t=10/1=10 m/s2 x2=v’t+at2/2=10x1+10x1/2 =10+5 =15 m v=dx/dt=8t-9t2; aavg=v/ t= =[v(t=2s)-v(t=1s)]/(2-1) = [-20-(-1)]/1=-19 m/s2

  3. Q4. Fig. 1 represents the velocity of a car (v) moving on a straight line as a function of time (t). Find the acceleration of the car at 6.0 s. (A ns: -3.0 m/s2) CH-2-072 a =v/t=(0-12)/(8-4)=-12/4= =-3m/s2

  4. CH-2-071 T071 : Q3.Fig 1 shows the position-time graph of an object. What is the average velocity of the object between t=0.0 s and t= 5.0 s? (Ans: 2.0 m/s) Q4. Fig 2 shows a velocity-time graph of a runner. If the runner starts from the origin, find his position at t = 4.0 s.( Ans: 45 m) x1= area of v-t graph x1= (1X10)/2+(1x10)+ (10x 2) + [(10x2)/2] xf-xi =5+10+20+10=45 m xf=45 m V= x/ t = xf-xi/tf-ti= = (10-0)/(5-0) =10/5=2 m/s

  5. CH-2-071 Q5.An object is thrown vertically upward with an initial speed of 25 m/s from the ground. What is the height of the object 1.0 s before it touches ground?(Ans:20 m) Q6.: A car starts from rest and accelerates at a rate of 2.0 m/s2 in a straight line until it reaches a speed of 20 m/s. The car then slows down at a constant rate of 1.0 m/s2 until it stops. How much time elapses (total time) from start to stop? (Ans: 30 s) Velocity at touch down =vf=-25 m/s Velocity 1s before vf=vi=vf+|g|t = -25+9.8x1=-15.2 m/s then y in last sec=(vf2-vi2)/(-2 |g|) y=[(-15)2 - (-15.2)2]/(-2x9.8) y=[(625-231)/(-19.6)=-19.99 m yi= - 20 m Total time= t1+t2 t1=vf-vi/a1=(20-0)/2 = 10 s t2= (0-20)/-1 =20 s Total time =10 s + 20 s= 30 s

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