diff options
Diffstat (limited to '3886/CH12')
30 files changed, 336 insertions, 0 deletions
diff --git a/3886/CH12/EX12.10/12_10.sce b/3886/CH12/EX12.10/12_10.sce new file mode 100644 index 000000000..73c10b698 --- /dev/null +++ b/3886/CH12/EX12.10/12_10.sce @@ -0,0 +1,14 @@ +//cage and mine shaft
+//t is time during which stone is in motion
+//s=(9.81*t^2)/2
+//consider motion of cage
+//t1 be the time taken to travel first 30 m
+a=0.6 //m/sec^2
+t1=10 //sec
+//When the stone strikes
+//s=(0.6*(t+10)^2)/2
+//solving
+t=3.286 //sec
+s=(9.81*3.286^2)/2 //m
+printf("\nt=%.2f sec\ns=%.2f m",t,s)
+
diff --git a/3886/CH12/EX12.10/12_10.txt b/3886/CH12/EX12.10/12_10.txt new file mode 100644 index 000000000..067c14e7a --- /dev/null +++ b/3886/CH12/EX12.10/12_10.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.10.sce', -1)
+
+t=3.29 sec
+s=52.96 m
\ No newline at end of file diff --git a/3886/CH12/EX12.11/12_11.sce b/3886/CH12/EX12.11/12_11.sce new file mode 100644 index 000000000..2344998d6 --- /dev/null +++ b/3886/CH12/EX12.11/12_11.sce @@ -0,0 +1,20 @@ +//Train B overtakes train A
+//refer fig.12.12
+//speed of A
+v1=7.5 //m/sec
+//speed of B
+v2=15 //m/sec
+//motion of train A
+//using equation of motion
+t1=7.5/0.15 //sec
+//distance travelled in time t
+//s=7.5*t-187.5
+//Motion of train B
+//using equation of motion
+t2=15/0.3 //sec
+//distance travelled t seconds after train A started
+//s=15*t-975
+//solving
+t=(975-187.5)/(15-7.5) //sec
+s=15*t-975 //m
+printf("\nTrain B overtakes train A %.2d sec\ns=%.2d m",t,s)
diff --git a/3886/CH12/EX12.11/12_11.txt b/3886/CH12/EX12.11/12_11.txt new file mode 100644 index 000000000..5a9698117 --- /dev/null +++ b/3886/CH12/EX12.11/12_11.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.11.sce', -1)
+
+Train B overtakes train A 105 sec
+s=600 m
\ No newline at end of file diff --git a/3886/CH12/EX12.12/12_12.sce b/3886/CH12/EX12.12/12_12.sce new file mode 100644 index 000000000..07c7e6818 --- /dev/null +++ b/3886/CH12/EX12.12/12_12.sce @@ -0,0 +1,29 @@ +//Two cars
+//refer fig.12.13
+//Let A and B be the positions of cars when the drivers see each other and apply brakes
+//Let they meet at C
+//1.car A
+au=12 //m/sec
+av=0
+//s=x
+//a1 be acceleration
+//using equation of motion
+//a1=(-12)/t
+//x=6*t
+//2.car B
+bu=9 //m/sec
+bv=0
+//a=a2
+//time=t
+//s=100-x
+//using equation of motion
+//a2=-9/t
+//100-x=4.5*t
+//solving
+t=100/10.5 //sec
+a1=-12/t //m/sec^2
+a2=-9/t //m/sec^2
+x=57.14 //m
+//distance traveled by second car
+bx=100-x //m
+printf("\nt=%.2f sec\na1=%.2f m/sec^2\na2=%.2f m/sec^2\nDistance travelled by first car=%.2f m\nDistance travelled by second car=%.2f m",t,a1,a2,x,bx)
diff --git a/3886/CH12/EX12.12/12_12.txt b/3886/CH12/EX12.12/12_12.txt new file mode 100644 index 000000000..e96634eef --- /dev/null +++ b/3886/CH12/EX12.12/12_12.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.12.sce', -1)
+
+t=9.52 sec
+a1=-1.26 m/sec^2
+a2=-0.95 m/sec^2
+Distance travelled by first car=57.14 m
+Distance travelled by second car=42.86 m
\ No newline at end of file diff --git a/3886/CH12/EX12.13/12_13.sce b/3886/CH12/EX12.13/12_13.sce new file mode 100644 index 000000000..f245fc9fb --- /dev/null +++ b/3886/CH12/EX12.13/12_13.sce @@ -0,0 +1,14 @@ +//Car and truck
+//refer fig.12.14 and 12.15
+u=12.5 //m/sec
+//sT=10+12.5*t+(aT*t^2)/2
+aT=-2 //m/sec^2
+//t is the time at any instant after the brakes are applied
+//sT=10+12.5*t-t^2
+//distance moved by car
+//sC=u*2+u*(t-2)+(aC*(t-2)^2)/2
+//sT=sC
+//Apply equations of motion
+//we get quadratic equation whose solution gives
+aC=-10/3 //m/sec^2
+printf("the deceleration of the car is=%.2f m/sec^2",aC)
diff --git a/3886/CH12/EX12.13/12_13.txt b/3886/CH12/EX12.13/12_13.txt new file mode 100644 index 000000000..06606ec3b --- /dev/null +++ b/3886/CH12/EX12.13/12_13.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.13.sce', -1)
+the deceleration of the car is=-3.33 m/sec^2
\ No newline at end of file diff --git a/3886/CH12/EX12.14/12_14.sce b/3886/CH12/EX12.14/12_14.sce new file mode 100644 index 000000000..5076245d2 --- /dev/null +++ b/3886/CH12/EX12.14/12_14.sce @@ -0,0 +1,13 @@ +//motion of particle
+//s=t^3-3*t^2+2*t+5
+//v=ds/dt=3*t^2-6*t+2
+//a=6*t-6
+//after 4 seconds
+v=3*4*4-6*4+2 //m/sec
+a=6*4-6 //m/sec^2
+//minimum velocity Vmin by using maxima and minima principle
+Vmin=-1 //m/sec
+//let at time t the velocity be zero,then
+t1=1.577 //sec
+t2=0.423 //sec
+printf("\nv=%.2f m/sec\na=%.2f m/sec^2\nMinimum velocity=%.2f m/sec\nVelocity is zero at t=%.2f sec and %.2f sec",v,a,Vmin,t1,t2)
diff --git a/3886/CH12/EX12.14/12_14.txt b/3886/CH12/EX12.14/12_14.txt new file mode 100644 index 000000000..2d8bf6721 --- /dev/null +++ b/3886/CH12/EX12.14/12_14.txt @@ -0,0 +1,7 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.14.sce', -1)
+
+v=26.00 m/sec
+a=18.00 m/sec^2
+Minimum velocity=-1.00 m/sec
+Velocity is zero at t=1.58 sec and 0.42 sec
\ No newline at end of file diff --git a/3886/CH12/EX12.15/12_15.sce b/3886/CH12/EX12.15/12_15.sce new file mode 100644 index 000000000..96c055e8a --- /dev/null +++ b/3886/CH12/EX12.15/12_15.sce @@ -0,0 +1,16 @@ +//particle in motion
+//v=t^3-t^2-2*t+2
+//a=3*t^2-2*t-2
+//acceleration after 4 seconds
+a=3*4^2-2*4-2 //m/sec^2
+//s=(t^4)/4-(t^3)/3-(t^2)+2*t+C
+//c is constant of acceleration
+//applying given condition
+C=4/3
+s=(4^4)/4-(4^3)/3-(4^2)+2*4+(4/3) //m
+//using maxima and minima principle
+//minimum value of acceleration (amin)
+amin=3*((1/3)^2)-2*(1/3)-2 //m/sec^2
+printf("\nMinimum value of acceleration=%.2f m/sec^2",amin)
+
+
diff --git a/3886/CH12/EX12.15/12_15.txt b/3886/CH12/EX12.15/12_15.txt new file mode 100644 index 000000000..d76f99291 --- /dev/null +++ b/3886/CH12/EX12.15/12_15.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.15.sce', -1)
+
+Minimum value of acceleration=-2.33 m/sec^2
\ No newline at end of file diff --git a/3886/CH12/EX12.16/12_16.sce b/3886/CH12/EX12.16/12_16.sce new file mode 100644 index 000000000..2c4775d4b --- /dev/null +++ b/3886/CH12/EX12.16/12_16.sce @@ -0,0 +1,22 @@ +//body moving along straight line
+//refer fig. 12.16
+//a=2-3*t
+//v=2*t-(3/2)*(t^2)+C1
+//C1 is constant of integration
+//v=20 //m/sec
+//t=5 //sec
+//thus
+C1=47.5
+//s=47.5*t+t^2-0.5*t^3+C2
+//s=85 m when t=10 sec thus
+C2=10
+a=2-3*0 //m/sec^2
+v=47.5 //m/sec
+as=10 //m
+//let the time when velocity becomes zero be t, thus
+t=6.33 //sec
+//Corresponding distance from origin
+s=10+47.5*6.33+6.33^2-0.5*6.33^3
+printf("\na=%.2f m/sec^2\nv=%.2f m/sec\ns=%.2f m\nt=%.2f sec\nDistance from origin\ns=%.3f",a,v,as,t,s)
+
+
diff --git a/3886/CH12/EX12.16/12_16.txt b/3886/CH12/EX12.16/12_16.txt new file mode 100644 index 000000000..60d5088f9 --- /dev/null +++ b/3886/CH12/EX12.16/12_16.txt @@ -0,0 +1,9 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.16.sce', -1)
+
+a=2.00 m/sec^2
+v=47.50 m/sec
+s=10.00 m
+t=6.33 sec
+Distance from origin
+s=223.926
\ No newline at end of file diff --git a/3886/CH12/EX12.18/12_18.sce b/3886/CH12/EX12.18/12_18.sce new file mode 100644 index 000000000..bc53c3a12 --- /dev/null +++ b/3886/CH12/EX12.18/12_18.sce @@ -0,0 +1,14 @@ +//Car moving
+//let the expression for retardation be
+//a=-k*s ...k=constant
+//v^2/2=(-k*(s^2)/2)+C1
+//When brakes are applied
+//s=0 and v=72 kmph
+v=20 //m/sec
+C1=200
+//when vehicle stops
+//v=0 s=15 m
+k=400/225
+//expression for retardation is
+//a=-1.778*s ...theory approach
+printf("The expression for retardation is a=-1.778*s")
\ No newline at end of file diff --git a/3886/CH12/EX12.18/12_18.txt b/3886/CH12/EX12.18/12_18.txt new file mode 100644 index 000000000..95c1bcb9d --- /dev/null +++ b/3886/CH12/EX12.18/12_18.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.18.sce', -1)
+The expression for retardation is a=-1.778*s
\ No newline at end of file diff --git a/3886/CH12/EX12.2/12_2.sce b/3886/CH12/EX12.2/12_2.sce new file mode 100644 index 000000000..9e1744a76 --- /dev/null +++ b/3886/CH12/EX12.2/12_2.sce @@ -0,0 +1,21 @@ +//Steel ball shot vertically up
+//refer fig.12.6
+//For upward motion
+au=18 //m/sec
+av=0
+aa=-9.81 //m/sec^2
+//s=h
+//let t1 be the time required to reach maximum height
+t1=1.83 //sec
+h=(18^2)/(2*9.81) //m
+//total height from the ground
+ah=25+h //m
+//Downward motion
+bu=0
+bs=41.51 //m
+ba=9.81 //m/sec^2
+v2=sqrt(2*9.81*41.51) //m/sec
+t2=28.54/9.81 //m/sec
+//total time during which the body is in motion
+t=t1+t2 //sec
+printf("\nt1=%.2f sec\nh=%.2f m\nv2=%.2f m/sec\nt2=%.2f sec\nt=%.2f sec",t1,h,v2,t2,t)
diff --git a/3886/CH12/EX12.2/12_2.txt b/3886/CH12/EX12.2/12_2.txt new file mode 100644 index 000000000..95719b60c --- /dev/null +++ b/3886/CH12/EX12.2/12_2.txt @@ -0,0 +1,8 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.2.sce', -1)
+
+t1=1.83 sec
+h=16.51 m
+v2=28.54 m/sec
+t2=2.91 sec
+t=4.74 sec
\ No newline at end of file diff --git a/3886/CH12/EX12.3/12_3.sce b/3886/CH12/EX12.3/12_3.sce new file mode 100644 index 000000000..2bcb11b87 --- /dev/null +++ b/3886/CH12/EX12.3/12_3.sce @@ -0,0 +1,13 @@ +//Height from which stone fell
+//refer fig.12.7
+//Let the stone be dropped from A at a height h above window
+//h=(g*t^2)/2 ...(1)
+//h+2.45=((g)*(t+0.5)^2)/2 ...(2)
+//from (1) and (2)
+t=0.2495 //sec
+g=9.81 //m/sec^2
+h=(g*t^2)/2 //m
+printf("\nh=%.3f m",h)
+
+
+
diff --git a/3886/CH12/EX12.3/12_3.txt b/3886/CH12/EX12.3/12_3.txt new file mode 100644 index 000000000..d7144aab7 --- /dev/null +++ b/3886/CH12/EX12.3/12_3.txt @@ -0,0 +1,4 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.3.sce', -1)
+
+h=0.305 m
\ No newline at end of file diff --git a/3886/CH12/EX12.4/12_4.sce b/3886/CH12/EX12.4/12_4.sce new file mode 100644 index 000000000..cc04c1e0f --- /dev/null +++ b/3886/CH12/EX12.4/12_4.sce @@ -0,0 +1,24 @@ +//Crossing of balls
+//refer fig. 12.8
+//1.for motion of first ball
+au=0
+//1s=30-h
+aa=9.81 //m/sec^2
+//2.for motion of second ball
+bu=15 //m/sec
+//s=h
+ba=-9.81 //m/sec^2
+//30-h=0*t+(9.81*t^2)/2 ...(1)
+//h=15*t-(9.81*t^2)/2 ...(2)
+//solving (1) and (2)
+t=30/15
+h=15*2-(9.81*2^2)/2 //m
+//at t=2
+//downward velocity of first ball
+v1=0+9.81*2 //m/sec
+//Upward velocity of second ball
+v2=15-9.81*2 //m/sec
+//relative velocity vr
+vr=v1-(-v2) //m/sec
+printf("\nt=%.2f sec\nh=%.2f m\nvr=%.2f m/sec",t,h,vr)
+
diff --git a/3886/CH12/EX12.4/12_4.txt b/3886/CH12/EX12.4/12_4.txt new file mode 100644 index 000000000..1d8a08fce --- /dev/null +++ b/3886/CH12/EX12.4/12_4.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.4.sce', -1)
+
+t=2.00 sec
+h=10.38 m
+vr=15.00 m/sec
\ No newline at end of file diff --git a/3886/CH12/EX12.5/12_5.sce b/3886/CH12/EX12.5/12_5.sce new file mode 100644 index 000000000..9a2ca3f70 --- /dev/null +++ b/3886/CH12/EX12.5/12_5.sce @@ -0,0 +1,13 @@ +//Stone dropped into well
+//let
+//h=depth of well
+//t1=time tataken by stone to strike water
+//t2=time taken by sound to travel h
+//t1+t2=4
+//h=(g*t1^2)/2
+//h=335*t2
+//solving
+//t1^2+68.30*t1-273.19=0
+t1=3.79 //sec
+h=(9.81*t1^2)/2 //m
+printf("h=%.2f m",h)
diff --git a/3886/CH12/EX12.5/12_5.txt b/3886/CH12/EX12.5/12_5.txt new file mode 100644 index 000000000..b50a1c27c --- /dev/null +++ b/3886/CH12/EX12.5/12_5.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.5.sce', -1)
+h=70.46 m
\ No newline at end of file diff --git a/3886/CH12/EX12.6/12_6.sce b/3886/CH12/EX12.6/12_6.sce new file mode 100644 index 000000000..bf9583733 --- /dev/null +++ b/3886/CH12/EX12.6/12_6.sce @@ -0,0 +1,19 @@ +//Distance covered
+//refer fig.12.9
+//Let the particle start from A and come to halt at E
+//Let initial velocity be u m/sec
+//consider motion between A and B
+//u+a=10
+//consider motion between A and C
+//70=7*u+7*a
+//solving
+a=-10/17.5 //m/sec^2
+u=10-(a) //m/sec
+//Let distance AD be s1
+s1=10.571*10+(-0.571*10^2)/2 //m
+//Distance covered in the interval 7 sec to 10 sec
+CD=77.16-60 //m
+//Let AE=s
+s=(10.571^2)/(2*0.571) //m
+printf("\nCD=%.2f m\ns=%.2f m",CD,s)
+
diff --git a/3886/CH12/EX12.6/12_6.txt b/3886/CH12/EX12.6/12_6.txt new file mode 100644 index 000000000..3115d1422 --- /dev/null +++ b/3886/CH12/EX12.6/12_6.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.6.sce', -1)
+
+CD=17.16 m
+s=97.85 m
\ No newline at end of file diff --git a/3886/CH12/EX12.7/12_7.sce b/3886/CH12/EX12.7/12_7.sce new file mode 100644 index 000000000..caca3452b --- /dev/null +++ b/3886/CH12/EX12.7/12_7.sce @@ -0,0 +1,13 @@ +//motorist and traffic light
+//initial velocity
+u=(80*1000)/(60*60) //m/sec
+t=10 //sec
+s=200 //m
+//a be acceleration
+//using equation of motion
+a=(200-22.22*10)*2/10^2 //m/sec^2
+//final velocity
+v=(22.22-0.444*10)*(3600/1000) //kmph
+printf("\na=%.2f m/sec^2\nv=%.2f kmph",a,v)
+
+
diff --git a/3886/CH12/EX12.7/12_7.txt b/3886/CH12/EX12.7/12_7.txt new file mode 100644 index 000000000..c1888fa21 --- /dev/null +++ b/3886/CH12/EX12.7/12_7.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.7.sce', -1)
+
+a=-0.44 m/sec^2
+v=64.01 kmph
\ No newline at end of file diff --git a/3886/CH12/EX12.9/12_9.sce b/3886/CH12/EX12.9/12_9.sce new file mode 100644 index 000000000..ea7da1ee6 --- /dev/null +++ b/3886/CH12/EX12.9/12_9.sce @@ -0,0 +1,13 @@ +//time required to cover the distance between two stations
+//refer fig.12.11
+v=(48*1000)/(60*60) //m/sec
+t1=30 //sec
+//after application of brakes the vehicle retards from 13.33 m/sec to 0 in t3 sec
+t3=13.33 //sec
+//Let t2 be the time during which the automobile travels with uniform velocity
+//s=s1+s2+s3
+s=5200 //m
+t2=((5200)-(13.33*30/2)-(13.33*13.33/2))/13.33 //sec
+//total time
+t=t1+t2+t3 //sec
+printf("Total time taken=%.2f sec",t)
diff --git a/3886/CH12/EX12.9/12_9.txt b/3886/CH12/EX12.9/12_9.txt new file mode 100644 index 000000000..ea732ee2b --- /dev/null +++ b/3886/CH12/EX12.9/12_9.txt @@ -0,0 +1,3 @@ +
+--> exec('E:\My program EM\12. Linear motion\12.9.sce', -1)
+Total time taken=411.76 sec
\ No newline at end of file |