From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001 From: priyanka Date: Wed, 24 Jun 2015 15:03:17 +0530 Subject: initial commit / add all books --- 3204/CH14/EX14.10/Ex14_10.sce | 14 ++++++++++++++ 3204/CH14/EX14.11/Ex14_11.sce | 17 +++++++++++++++++ 3204/CH14/EX14.12/Ex14_12.sce | 13 +++++++++++++ 3204/CH14/EX14.15/Ex14_15.sce | 9 +++++++++ 3204/CH14/EX14.16/Ex14_16.sce | 13 +++++++++++++ 3204/CH14/EX14.17/Ex14_17.sce | 20 ++++++++++++++++++++ 3204/CH14/EX14.18/Ex14_18.sce | 12 ++++++++++++ 3204/CH14/EX14.19/Ex14_19.sce | 12 ++++++++++++ 3204/CH14/EX14.20/Ex14_20.sce | 13 +++++++++++++ 3204/CH14/EX14.21/Ex14_21.sce | 21 +++++++++++++++++++++ 3204/CH14/EX14.22/Ex14_22.sce | 13 +++++++++++++ 3204/CH14/EX14.23/Ex14_23.sce | 10 ++++++++++ 3204/CH14/EX14.24/Ex14_24.sce | 13 +++++++++++++ 3204/CH14/EX14.25/Ex14_25.sce | 22 ++++++++++++++++++++++ 3204/CH14/EX14.3/Ex14_3.sce | 15 +++++++++++++++ 3204/CH14/EX14.30/Ex14_30.sce | 30 ++++++++++++++++++++++++++++++ 3204/CH14/EX14.4/Ex14_4.sce | 16 ++++++++++++++++ 3204/CH14/EX14.5/Ex14_5.sce | 12 ++++++++++++ 3204/CH14/EX14.6/Ex14_6.sce | 12 ++++++++++++ 3204/CH14/EX14.7/Ex14_7.sce | 22 ++++++++++++++++++++++ 3204/CH14/EX14.9/Ex14_9.sce | 16 ++++++++++++++++ 21 files changed, 325 insertions(+) create mode 100644 3204/CH14/EX14.10/Ex14_10.sce create mode 100644 3204/CH14/EX14.11/Ex14_11.sce create mode 100644 3204/CH14/EX14.12/Ex14_12.sce create mode 100644 3204/CH14/EX14.15/Ex14_15.sce create mode 100644 3204/CH14/EX14.16/Ex14_16.sce create mode 100644 3204/CH14/EX14.17/Ex14_17.sce create mode 100644 3204/CH14/EX14.18/Ex14_18.sce create mode 100644 3204/CH14/EX14.19/Ex14_19.sce create mode 100644 3204/CH14/EX14.20/Ex14_20.sce create mode 100644 3204/CH14/EX14.21/Ex14_21.sce create mode 100644 3204/CH14/EX14.22/Ex14_22.sce create mode 100644 3204/CH14/EX14.23/Ex14_23.sce create mode 100644 3204/CH14/EX14.24/Ex14_24.sce create mode 100644 3204/CH14/EX14.25/Ex14_25.sce create mode 100644 3204/CH14/EX14.3/Ex14_3.sce create mode 100644 3204/CH14/EX14.30/Ex14_30.sce create mode 100644 3204/CH14/EX14.4/Ex14_4.sce create mode 100644 3204/CH14/EX14.5/Ex14_5.sce create mode 100644 3204/CH14/EX14.6/Ex14_6.sce create mode 100644 3204/CH14/EX14.7/Ex14_7.sce create mode 100644 3204/CH14/EX14.9/Ex14_9.sce (limited to '3204/CH14') diff --git a/3204/CH14/EX14.10/Ex14_10.sce b/3204/CH14/EX14.10/Ex14_10.sce new file mode 100644 index 000000000..b8dd45d23 --- /dev/null +++ b/3204/CH14/EX14.10/Ex14_10.sce @@ -0,0 +1,14 @@ +// Initilization of variables +a=10 // m/s^2 // acceleration of the particle +S_5th=50 // m // distance travelled by the particle during the 5th second +t=5 // seconds +// Calculations +// The distance travelled by the particle in time t is given by, S=(u*t)+(1/2)*a*t^2.....(consider this as eq'n 1) +// Here, The distance travelled by the particle in the 5th second=The distance travelled in 5 seconds - The distance travelled in 4 seconds..... (consider eq'n 2) +// Using eq'n 1: S_(0-5)=(5*u)+(1/2)*10*5^2 = 5*u+125.....(consider eq'n 3) +// again, S_(0-4)=(4*u)+(1/2)*10*4^2 = 4*u+80....(consider eq'n 4) +// Now,put eq'n 3&4 in eq'n 2 and solve for u. We get, 50=[(5*u+125)-(4*u+80)] i.e 50=u+45 +u=(S_5th)-45 // m/s +// Calculations +clc +printf('The initial velocity of the particle is %f m/s \n',u) diff --git a/3204/CH14/EX14.11/Ex14_11.sce b/3204/CH14/EX14.11/Ex14_11.sce new file mode 100644 index 000000000..38ff8a24c --- /dev/null +++ b/3204/CH14/EX14.11/Ex14_11.sce @@ -0,0 +1,17 @@ +// Initilization of variables +// Conditions given are +t=1 // s +x=14.75 // m +v=6.33 // m/s +// Calculations +// We use expression 1,2 & 3 to find distance,velocity & acceleration of the particle after 2 sec +T=2 // sec +X=(T^4/12)-(T^3/3)+(T^2)+(5*T)+9 // m // eq'n 3 +V=(T^3/3)-(T^2)+(2*T)+5 // m/s +a=(T^2)-(2*T)+2 // m/s^2 +// Results +clc +printf('The distance travelled by the particle is %f m \n',X) +printf('The velocity of the particle is %f m/s \n',V) +printf('The acceleration of the particle is %f m/s^2 \n',a) +// The answer may vary due to decimal point error diff --git a/3204/CH14/EX14.12/Ex14_12.sce b/3204/CH14/EX14.12/Ex14_12.sce new file mode 100644 index 000000000..fdf68968f --- /dev/null +++ b/3204/CH14/EX14.12/Ex14_12.sce @@ -0,0 +1,13 @@ +// Calculations +// From eq'n 2 it is clear that velocity of the particle becomes zero at t=3 sec +t=3 // sec .. from eq'n 2 +// Position of particle at t=3 sec +x=(t^3)-(3*t^2)-(9*t)+12 // m // from eq'n 1 +// Acc of particle at t=3 sec +a=6*(t-1) // m/s^2 // from eq'n 3 +// Results +clc +printf('The time at which the velocity of the particle becomes zero is %f sec \n',t) +printf('The position of the partice at t=3 sec is %f m \n',x) +printf('The acceleration of the particle is %f m/s^2 \n',a) +// Ref textbook for the graphs diff --git a/3204/CH14/EX14.15/Ex14_15.sce b/3204/CH14/EX14.15/Ex14_15.sce new file mode 100644 index 000000000..42e8aac3a --- /dev/null +++ b/3204/CH14/EX14.15/Ex14_15.sce @@ -0,0 +1,9 @@ +// Initilization of variables +F=250 // N // Force acting on a body +m=100 // kg // mass of the body +// Calculations +// Using the eq'n of motion +a=F/m // m/s^2 +// Results +clc +printf('The acceleration of the body is %f m/s^2 \n',a) diff --git a/3204/CH14/EX14.16/Ex14_16.sce b/3204/CH14/EX14.16/Ex14_16.sce new file mode 100644 index 000000000..c3024a6c0 --- /dev/null +++ b/3204/CH14/EX14.16/Ex14_16.sce @@ -0,0 +1,13 @@ +// Initilization of variables +a=1 // m/s^2 // downward/upward acceleration of the elevator +W=500 // N // Weight of man +g=9.81 // m/s^2 // acceleration due to gravity +// Calculations +// (a) Downward Motion +R_1=W*(1-(a/g)) // N // (Assume pressure as R_1) +// (b) Upward Motion +R_2=W*(1+(a/g)) // N // (Assume pressure as R_2) +// Results +clc +printf('(a) The pressure transmitted to the floor by the man for Downward motion of the elevator is %f N \n',R_1) +printf('(b) The pressure transmitted to the floor by the man for Upward motion of the elevator is %f N \n',R_2) diff --git a/3204/CH14/EX14.17/Ex14_17.sce b/3204/CH14/EX14.17/Ex14_17.sce new file mode 100644 index 000000000..f6e47e71e --- /dev/null +++ b/3204/CH14/EX14.17/Ex14_17.sce @@ -0,0 +1,20 @@ +// Initilization of variables +W=5000 // N // Total weight of the elevator +u=0 // m/s +v=2 // m/s // velocity of the elevator +s=2 // m // distance traveled by the elevator +t=2 // seconds // time to stop the lift +w=600 // N // weight of the man +g=9.81 // m/s^2 // acc due to gravity +// Calculations +// Acceleration acquired by the elevator after travelling 2 m is given by, +a=sqrt((v^2-u^2)/(2*s)) // m/s^2 +// (a) Let T be the the tension in the cable which is given by eq'n, +T=W*(1+(a/g)) // N +// (b) Motion of man +// Let R be the pressure experinced by the man.Then from the Eq'n of motion of man pressure is given as, +R=w*(1-(a/g)) // N +// Results +clc +printf('(a) The Tensile force in the cable is %f N \n',T) +printf('(b) The pressure transmitted to the floor by the man is %f N \n',R) diff --git a/3204/CH14/EX14.18/Ex14_18.sce b/3204/CH14/EX14.18/Ex14_18.sce new file mode 100644 index 000000000..40bbf8e27 --- /dev/null +++ b/3204/CH14/EX14.18/Ex14_18.sce @@ -0,0 +1,12 @@ +// Initilization of variables +M_1=10 // kg // mass of the 1st block +M_2=5 // kg // mass of the 2nd block +mu=0.25 // coefficient of friction between the blocks and the surface +g=9.81 // m/s^2 // acc due to gravity +// Calculations +a=g*(M_2-(mu*M_1))/(M_1+M_2) // m/s^2 // from eq'n 5 +T=M_1*M_2*g*(1+mu)/(M_1+M_2) // N // from eq'n 6 +// Results +clc +printf('The acceleration of the masses is %f m/s^2 \n',a) +printf('The tension in the string is %f N \n',T) diff --git a/3204/CH14/EX14.19/Ex14_19.sce b/3204/CH14/EX14.19/Ex14_19.sce new file mode 100644 index 000000000..9218ea01d --- /dev/null +++ b/3204/CH14/EX14.19/Ex14_19.sce @@ -0,0 +1,12 @@ +// Initilization of variables +M_1=150 // kg // mass of the 1st block +M_2=100 // kg // mass of the 2nd block +mu=0.2 // coefficient of friction between the blocks and the inclined plane +g=9.81 // m/s^2 // acc due to gravity +theta=45 // degree // inclination of the surface +// Calculations +// substuting the value of eq'n 3 in eq'n 1 & solving for T,we get value of T as, +T=((M_1*M_2*g)*(sind(theta)+2-(mu*cosd(theta))))/((4*M_1)+(M_2)) // N +// Results +clc +printf('The tension in the string during the motion of the system is %f N \n',T) diff --git a/3204/CH14/EX14.20/Ex14_20.sce b/3204/CH14/EX14.20/Ex14_20.sce new file mode 100644 index 000000000..abb0e4025 --- /dev/null +++ b/3204/CH14/EX14.20/Ex14_20.sce @@ -0,0 +1,13 @@ +// Initilization of variables +M_1=5 // kg // mass of the 1st block +theta_1=30 // degree // inclination of the 1st plane +M_2=10 // kg // mass of the 2nd block +theta_2=60 // degree // inclination of the 2nd plane +mu=0.33 // coefficient of friction between the blocks and the inclined plane +g=9.81 // m/s^2 // acc due to gravity +// Calculations +// solving eq'n 1 & 2 for a we get, +a=((((M_2*(sind(theta_2)-(mu*cosd(theta_2))))-(M_1*(sind(theta_1)+(mu*cosd(theta_1))))))*g)/(M_1+M_2) // m/s^2 +// Results +clc +printf('The acceleration of the masses is %f m/s^2 \n',a) diff --git a/3204/CH14/EX14.21/Ex14_21.sce b/3204/CH14/EX14.21/Ex14_21.sce new file mode 100644 index 000000000..f6aa0ef95 --- /dev/null +++ b/3204/CH14/EX14.21/Ex14_21.sce @@ -0,0 +1,21 @@ +// Initilization of variables +S=5 // m // distance between block A&B +mu_A=0.2 // coefficient of friction between the block A and the inclined plane +mu_B=0.1 // coefficient of friction between the block B and the inclined plane +theta=20 // degree // inclination of the pane +g=9.81 // m/s^2 // acc due to gravity +// Calculatio// +// EQUATION OF MOTION OF BLOCK A: +// Let a_A & a_B be the acceleration of block A & B. +a_A=(g*sind(theta))-(mu_A*g*cosd(theta)) // m/s^2 // from eq'n 1 & eq'n 2 +// EQUATION OF MOTION OF BLOCK B: +a_B=g*((sind(theta))-(mu_B*cosd(theta))) // m/s^2 // from eq'n 3 & Rb +// Now the eq'n for time of collision of the blocks is given as, +t=sqrt((S*2)/(a_B-a_A)) // seconds +S_A=(1/2)*a_A*t^2 // m // distance travelled by block A +S_B=(1/2)*a_B*t^2 // m // distance travelled by block B +// Results +clc +printf('The time before collision is %f seconds \n',t) +printf('The distance travelled by block A before collision is %f m \n',S_A) +printf('The distance travelled by block B before collision is %f m \n',S_B) diff --git a/3204/CH14/EX14.22/Ex14_22.sce b/3204/CH14/EX14.22/Ex14_22.sce new file mode 100644 index 000000000..8aad11eba --- /dev/null +++ b/3204/CH14/EX14.22/Ex14_22.sce @@ -0,0 +1,13 @@ +// Initilization of variables +P=50 // N // Weight of the car +Q=100 // N // Weight of the rectangular block +g=9.81 // m/s^2 // acc due to gravity +b=25 // cm // width of the rectangular block +d=50 // cm // depth of the block +// Calculations +a=(Q*g)/(4*P+2*Q) // m/s^2 // from eq'n 4 +W=(Q*(P+Q))/(4*P+Q) // N // from eq'n 6 +// Resuts +clc +printf('The maximum value of weight (W) by which the car can be accelerated is %f N \n',W) +printf('The acceleration is %f m/s^2 \n',a) diff --git a/3204/CH14/EX14.23/Ex14_23.sce b/3204/CH14/EX14.23/Ex14_23.sce new file mode 100644 index 000000000..95fcf24f1 --- /dev/null +++ b/3204/CH14/EX14.23/Ex14_23.sce @@ -0,0 +1,10 @@ +// Initilization of variables +P=40 // N // weight on puley r_1 +Q=60 // N // weight on pulley r_2 +g=9.81 // m/s^2 // acc due to gravity +// Calculations +// The eq'n for acceleration of pulley Pi.e a_p is, +a_p=(((2*P)-(Q))/((4*P)+(Q)))*2*g // m/s^2 +// Results +clc +printf('The downward acceleration of P is %f m/s^2 \n',a_p) diff --git a/3204/CH14/EX14.24/Ex14_24.sce b/3204/CH14/EX14.24/Ex14_24.sce new file mode 100644 index 000000000..a14a9e70a --- /dev/null +++ b/3204/CH14/EX14.24/Ex14_24.sce @@ -0,0 +1,13 @@ +// Initilization of variables +M=15 // kg // mass of the wedge +m=6 // kg // mass of the block +theta=30 // degree // angle of the wedge +g=9.81 // m/s^2 // acc due to gravity +// Calculations +a_A=((m*g*cosd(theta)*sind(theta))/((M)+(m*(sind(theta))^2)))/(g) // g // By eliminating R_1 from eq'n 1&3. +// Here, assume a_r is the acceleration of block B relative to wedge A which is given by substuting a_A in eq'n 2 +a_r=(((g*sind(theta))*(m+M))/((M)+(m*(sind(theta))^2)))/(g) // g +// Results +clc +printf('(a) The acceleration of the wedge is %f g \n',a_A) +printf('(b) The acceleration of the bock relative to the wedge is %f g \n',a_r) diff --git a/3204/CH14/EX14.25/Ex14_25.sce b/3204/CH14/EX14.25/Ex14_25.sce new file mode 100644 index 000000000..a3404c0bd --- /dev/null +++ b/3204/CH14/EX14.25/Ex14_25.sce @@ -0,0 +1,22 @@ +// Initilization of variables +P=30 // N // weight on pulley A +Q=20 // N // weight on pulley B +R=10 // N // weight on puey B +g=9.81 // m/s^2 // acc due to gravity +// Calculations +// Solving eqn's 6 & 7 using matrix for a & a_1, we get +A=[70 -40;-10 30] +B=[10;-10] +C=inv(A)*B +// Acceleration of P is given as, +P=C(1) // m/s^2 +// Acceleration of Q is given as, +Q=C(2)-C(1) // m/s^2 +// Acceleration of R is given as, +R=-(C(2)+C(1)) // m/s^2 // as R is taken to be +ve +// Results +clc +printf('The acceleration of P is %f g \n',P) +printf('The acceleration of Q is %f g \n',Q) +printf('The acceleration of R is %f g \n',R) +// Here the -ve sign indicates deceleration or backward/downward acceleation. diff --git a/3204/CH14/EX14.3/Ex14_3.sce b/3204/CH14/EX14.3/Ex14_3.sce new file mode 100644 index 000000000..9db304e2c --- /dev/null +++ b/3204/CH14/EX14.3/Ex14_3.sce @@ -0,0 +1,15 @@ +// Initilization of variables +a_T=0.18 // m/s^2 // acc of trolley +// Calculations +a_B=-a_T/3 // m/s^2 // from eq'n 4 +t=4 // seconds +v_T=a_T*t // m/s // velocity of trolley after 4 seconds +v_B=-v_T/3 // m/s // from eq'n 3 +S_T=(1/2)*a_T*t^2 // m // distance moved by trolley in 4 sec +S_B=-S_T/3 // m // from eq'n 2 +// Results +clc +printf('The acceleration of block B is %f m/s^2 \n',a_B) +printf('The velocity of trolley & the block after 4 sec is %f m/s & %f m/s \n',v_T,v_B) +printf('The distance moved by the trolley & the block is %f m & %f m \n',S_T,S_B) +// The -ve sign indicates that the velocity or the distance travelled is in opposite direction. diff --git a/3204/CH14/EX14.30/Ex14_30.sce b/3204/CH14/EX14.30/Ex14_30.sce new file mode 100644 index 000000000..1453cbdb0 --- /dev/null +++ b/3204/CH14/EX14.30/Ex14_30.sce @@ -0,0 +1,30 @@ +// Initilization of variables +W=1 // kg/m // weight of the bar +L_AB=0.6 // m // length of segment AB +L_BC=0.30 // m // length of segment BC +g=9.81 // m/s^2 // acc due to gravity +// Calculations +// Consider the respective F.B.D. +theta_1=atand(5/12) // slope of bar AB // here theta_1= atan(theta) +theta_2=asind(5/13) // theta_2=asin(theta) +theta_3=acosd(12/13) // theta_3=acos(theta) +M_AB=L_AB*W // kg acting at D // Mass of segment AB +M_BC=L_BC*W // kg acting at E // Mass of segment BC +// The various forces acting on the bar are: +// Writing the eqn's of dynamic equilibrium +Y_A=(L_AB*g)+(L_BC*g) // N // sum F_y=0 +// Using moment eq'n Sum M_A=0:Here,in this eq'n the values are as follows, +AF=L_BC*cosd(theta_3) +DF=L_BC*sind(theta_2) +AH=(L_AB*cosd(theta_3))+((L_BC/2)*sind(theta_2)) +IG=(L_AB*sind(theta_2))-((L_BC/2)*cosd(theta_3)) +// On simplifying and solving moment eq'n we get a as, +a=((2*L_AB*L_BC*g*sind(theta_2))-(L_BC*g*(L_BC/2)*cosd(theta_3)))/((2*L_AB*L_BC*cosd(theta_3))+(L_BC*(L_BC/2)*sind(theta_2))) // m/s^2 +X_A=0.9*a //N // from eq'n of dynamic equilibrium +R_A=sqrt(X_A^2+Y_A^2) // N // Resultant of R_A +alpha=atand(Y_A/X_A) // degree +// Results +clc +printf('The acceleration is %f m/s^2 \n',a) +printf('The reaction at A (R_A) is %f N \n',R_A) +printf('The angle made by the resultant is %f degree \n',alpha) diff --git a/3204/CH14/EX14.4/Ex14_4.sce b/3204/CH14/EX14.4/Ex14_4.sce new file mode 100644 index 000000000..602f17a8f --- /dev/null +++ b/3204/CH14/EX14.4/Ex14_4.sce @@ -0,0 +1,16 @@ +// Initiliztion of variables +v_B=12 // cm/s // velocity of block B +u=0 +s=24 // cm // distance travelled by bock B +t=5 // seconds +// Calculations +a_B=v_B^2/(2*s) // cm/s^2 // using eq'n v^2-u^2=28*a*s for block B. Here u=0 +a_A=(3/2)*a_B // cm/s^2 // from eq'n 4 // Here a_A is negative which means acceleration is in opposite direction. However we consider +ve values for further calculations +v_A=u+(a_A*t) // m/s // using eq'n v=u+(a*t) +S_A=(u*t)+((1/2)*a_A*t^2) // m // using eq'n S=(u*t)+((1/2)*a*t^2) +// Results +clc +printf('The acceleration of block A (a_A) is %f cm/s^2 \n',a_A) +printf('The acceleration of block B (a_B) is %f cm/s^2 \n',a_B) +printf('The velocity of block A (v_A) after 5 seconds is %f m/s \n',v_A) +printf('The position of block A (S_A) after 5 seconds is %f m \n',S_A) diff --git a/3204/CH14/EX14.5/Ex14_5.sce b/3204/CH14/EX14.5/Ex14_5.sce new file mode 100644 index 000000000..90aedae9d --- /dev/null +++ b/3204/CH14/EX14.5/Ex14_5.sce @@ -0,0 +1,12 @@ +// Initilization of variables +u=72*(1000/(60*60)) // km/hr // speed of the vehicle +s=300 // m // distance where the light is turning is red +t=20 // s // traffic light timed to remain red +// Calculations +// Now to find the acceleration we use the eq'n s=u*t+(1/2)*a*t^2 +a=(((s)-(u*t))*2)/t^2 // m/s^2 (Deceleration) +v=(u+(a*t))*((60*60)/1000) // km/hr // here we multiply with (60*60)/1000 to convert m/s to km/hr +// Results +clc +printf('(a) The required uniform acceleration of the car is %f m/s^2 \n',a) +printf('(b) The speed at which the motorist crosses the traffic light is %f km/hr \n',v) diff --git a/3204/CH14/EX14.6/Ex14_6.sce b/3204/CH14/EX14.6/Ex14_6.sce new file mode 100644 index 000000000..7bc6b29c7 --- /dev/null +++ b/3204/CH14/EX14.6/Ex14_6.sce @@ -0,0 +1,12 @@ +// Initilization of variables +S=50 // m // height of the tower +v=25 // m/s // velocity at which the stone is thrown up from the foot of the tower +g=9.81 // m/s^2 // acc due to graity +// Calculations +// The equation of time for the two stones to cross each other is given as, +t=S/v // seconds +S_1=(1/2)*g*t^2 // m // from the top +// Results +clc +printf('The time (t) at which the two stones cross each other is %f seconds \n',t) +printf('The two stones cross each other (from top) at a distance of %f m \n',S_1) diff --git a/3204/CH14/EX14.7/Ex14_7.sce b/3204/CH14/EX14.7/Ex14_7.sce new file mode 100644 index 000000000..5092b6be2 --- /dev/null +++ b/3204/CH14/EX14.7/Ex14_7.sce @@ -0,0 +1,22 @@ +// Intilization of variables +acc=0.5 // m/s^2 // acceleration of the elevator +s=25 // m // distance travelled by the elevator from the top +u=0 // m/s +g=9.81 // m/s^2 // acc due to gravity +// Calculations +// Using eq'n the eq'n v^2-u^2=2*a*s, solving for v we get, +v=sqrt((2*acc*s)+(u^2)) // m/s +// Now solving eq'n 1 & 2 for t we get, (4.655*t^2)-(5*t)+(25)=0 +// Find the roots of the eq'n using the eq'n,t=(-b+sqrt(b^2-(4*a*c)))/(2*a).In this eq'n the values of a,b & c are, +a=4.655 +b=-5 +c=-25 +t=(-b+sqrt((b^2)-(4*a*c)))/(2*a) // seconds +// Let S_1 be the distance travelled by the elevator after it travels 25 m from top when the stone hits the elevator,This disance S_1 is given as, +S_1=(v*t)+((1/2)*acc*t^2) // m +// Let S be the total dist from top when the stone hits the elevator, +S=S_1+s // m +// Results +clc +printf('The time taken by the stone to hit the elevator is %f seconds \n',t) +printf('The distance (S)travelled by the elevator at the time of impact is %f m \n',S) diff --git a/3204/CH14/EX14.9/Ex14_9.sce b/3204/CH14/EX14.9/Ex14_9.sce new file mode 100644 index 000000000..5fc24cb7e --- /dev/null +++ b/3204/CH14/EX14.9/Ex14_9.sce @@ -0,0 +1,16 @@ +// Initilization of variables +v=60 // km/hr // velocity of the train +d1=15 // km // Distance travelled by the local train from the velocity-time graph (here d1= Area OED) +d2=12 // km // from the velocity-time graph (here d2= Area OABB') +d3=3 // km // from the velocity-time graph (here d3= Area BB'C) +// Calculations +t_1=d2/v // hr // time of travel for first 12 kms +t_2=(2*d3)/v // hr // time of for next 3 kms +// Total time of travel for passenger train is given by eq'n +t=t_1+t_2 // hr +// Now time of travel of the local train (let it be T) is given as, +T=2*t // hr +V_max=(2*d1)/T // km/hr +// Results +clc +printf('The maximum speed of the local train is %f km/hr \n',V_max) -- cgit