Ch-2 Help-Session. CH-2-072 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.

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Presentation transcript:

Ch-2 Help-Session

CH 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) 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) v=dx/dt=8t-9t 2 ; a avg =  v/  t= =[v(t=2s)-v(t=1s)]/(2-1) = [-20-(-1)]/1=-19 m/s 2  x 1 =v i t+at 2 /2 v avg = (v i +v f )/2= v f /2=  x/  t = v f /2=  x/  t=5; v f =2x5=10 m/s a =v f -v i /t=10/1=10 m/s2  x 2 =v’t+at2/2=10x1+10x1/2 =10+5 =15 m

CH 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: m/s2 ) a =  v/  t=(0-12)/(8-4)=-12/4= =-3m/s 2

CH 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) V=  x/  t = x f -x i /t f -t i = = (10-0)/(5-0) =10/5=2 m/s  x 1 = area of v-t graph  x 1 = (1X10)/2+(1x10)+ (10x 2) + [(10x2)/2] x f -xi = =45 m x f =45 m

CH 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 = x1=-15.2 m/s then  y in last sec=( v f 2 -v i 2 )/(-2 |g|)  y=[(-15) 2 - (-15.2) 2 ]/(-2x9.8)  y=[( )/(-19.6)= m y i = - 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